Kinetic energy exercises

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Test your knowledge with questions about kinetic energy and solve your doubts with the commented resolution.

question 1

Calculate the kinetic energy of a ball of mass 0.6 kg as it is thrown and reaches a velocity of 5 m/s.

Correct answer: 7.5 J.

Kinetic energy is associated with the movement of a body and can be calculated using the following formula:

straight E with straight c subscript space equal to numerator space straight m space. straight space V squared over denominator 2 end of fraction

Substituting the question data in the formula above, we find the kinetic energy.

straight E with straight c subscript space equal to space numerator 0 comma 6 space kg space. space left parenthesis 5 straight space m divided by straight space s right parenthesis squared over denominator 2 end of fraction straight E with straight c subscript space equal to space numerator 0 comma 6 space kg space. space 25 straight space m squared divided by straight s squared over denominator 2 end of fraction straight E with straight c subscript space equal to 15 over 2 numerator kg space. straight space m squared over straight denominator s squared end of fraction straight E with straight c subscript space equal to space 7 comma 5 numerator kg space. straight space m squared over straight denominator s squared end of fraction equal to 7 comma 5 straight space J

Therefore, the kinetic energy acquired by the body during movement is 7.5 J.

question 2

A doll with a mass of 0.5 kg was dropped from a window on the 3rd floor, at a height of 10 m from the ground. What is the kinetic energy of the doll when it hits the ground and how fast did it fall? Consider the acceleration of gravity to be 10 m/s2.

Correct answer: kinetic energy of 50 J and speed of 14.14 m/s.

When playing the doll, work was done to move it and energy was transferred to it through movement.

The kinetic energy acquired by the doll during launch can be calculated by the following formula:

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straight delta space equal to straight space F. straight d straight delta space equal to straight space m. straight to. straight from

Replacing the utterance values, the kinetic energy resulting from the movement is:

straight delta space equal to space 0 comma 5 space kg space. space 10 straight space m divided by straight s squared space. space 10 space straight m straight delta space equal to 50 space numerator kg space. straight space m squared over straight denominator s squared end of fraction equal to space 50 straight space J

Using the other formula for kinetic energy, we calculate how fast the doll fell.

straight E with straight c subscript space equal to numerator space straight m space. straight space V squared over denominator 2 end of fraction 50 numerator space kg. straight m squared over denominator straight s squared end of fraction space equals space numerator 0 comma 5 space kg space. straight space V squared over denominator 2 end of fraction straight V squared space equal to space numerator 2 straight space x space 50 numerator kg. straight m squared over denominator straight s squared end of fraction over denominator 0 comma 5 space Kg end of fraction straight V squared space equal to numerator space 100 numerator space diagonal up risk kg. straight m squared over denominator straight s squared end of fraction over denominator 0 comma 5 diagonal space up risk Kg end of fraction straight V squared space equal to 200 straight space m squared divided by straight s squared straight V space equal to space squared root of 200 straight space m squared divided by straight s squared end of root straight V approximately equal space 14 comma 14 straight space m divided by straight only

Thus, the kinetic energy of the doll is 50 J and the speed it reaches is 14.14 m/s.

question 3

Determine the work done by a body of mass 30 kg so that its kinetic energy increases as its velocity increases from 5 m/s to 25 m/s?

Correct answer: 9000 J.

Work can be calculated by varying kinetic energy.

straight T space equal to space increment straight E with straight c subscript straight T space equal to space straight E with cf subscript space end of subscript minus straight space E with ci straight subscript T space equal to straight numerator m space. straight space V with straight f subscript with 2 superscript over denominator 2 end of fraction space minus space straight numerator m space. straight space V with straight i subscript with 2 superscript over denominator 2 end of fraction straight T space equal to straight m over 2. open parentheses straight V with straight f subscript with 2 superscript space minus straight space V with straight i subscript with 2 superscript close parentheses

Replacing the values ​​of the statement in the formula, we have:

straight T space equal to space numerator 30 space kg over denominator 2 end of fraction. space open brackets open parentheses 25 straight space m divided by straight s close square brackets space less space open parentheses 5 straight space m divided by straight s closes squared parentheses closes square brackets T space equal to 15 space kg space. space left parenthesis 625 straight space m squared divided by straight s squared space minus space 25 straight space m squared divided by straight s squared right parenthesis straight T space equal to 15 kg space space. space 600 straight space m squared divided by straight s squared straight T narrow space equal to space 9000 numerator space kg. straight m squared over straight denominator s squared end of fraction equal to space 9000 straight space J

Therefore, the work required to change the body's speed will be equal to 9000 J.

See too: Work

question 4

A motorcyclist is riding his motorcycle on a road with radar at a speed of 72 km/h. After passing through the radar, it accelerates and its speed reaches 108 km/h. Knowing that the mass of the motorcycle and rider combination is 400 kg, determine the variation in kinetic energy suffered by the rider.

Correct answer: 100 kJ.

We must first perform the conversion of the given speeds from km/h to m/s.

numerator 72 space km divided by straight h over denominator space 3 comma 6 end of fraction equal to space 20 straight space m divided by straight s
numerator 108 space km divided by straight h over denominator space 3 comma 6 end of fraction equal to space 30 straight space m divided by straight s

The change in kinetic energy is calculated using the formula below.

straight increment E with straight c subscript space equal to straight space E with cf subscript space end of subscript minus straight space E with ci subscript straight increment E with straight c subscript space equal to straight numerator m space. straight space V with straight f subscript with 2 superscript over denominator 2 end of fraction space minus space straight numerator m space. straight space V with straight i subscript with 2 superscript over denominator 2 end of fraction increment straight E with straight c subscript space equal to straight m over 2. open parentheses straight V with straight f subscript with 2 superscript space minus straight space V with straight i subscript with 2 superscript close parentheses

Substituting the problem values ​​in the formula, we have:

straight increment E with straight c subscript space equal to numerator 400 space kg over denominator 2 end of fraction. space open brackets open parentheses 30 straight space m divided by straight s close square brackets space less open parentheses 20 space straight m divided by straight s closes square brackets closes squares increment straight E with straight c subscript space equal to 200 space kg space. space opens parentheses 900 straight space m squared divided by straight s squared space minus space 400 straight space m squared square divided by straight s squared close parentheses straight increment E with straight c subscript space equal to 200 space kg space. space 500 straight space m squared divided by straight s squared increment straight E with straight c subscript space equal to 100 space 000 space numerator kg space. straight space m squared over straight denominator s squared end of fraction straight increment E with straight c subscript space equal to 100 space 000 straight space J space equal to space 100 space kJ

Thus, the kinetic energy variation in the path was 100 kJ.

question 5

(UFSM) A mass bus m travels along a mountain road and descends a height h. The driver keeps the brakes on so that the speed is kept constant in module throughout the journey. Consider the following statements, check whether they are true (T) or false (F).

( ) The kinetic energy variation of the bus is null.
( ) The mechanical energy of the bus-Earth system is conserved, as the speed of the bus is constant.
( ) The total energy of the bus-Earth system is conserved, although part of the mechanical energy is transformed into internal energy. The correct sequence is

a) V – F – F.
b) V – F – V.
c) F – F – V.
d) F – V – V.
e) F - V - F

Correct alternative: b) V – F – V.

(TRUE) The bus's kinetic energy variation is zero, as the speed is constant and the kinetic energy variation depends on changes in this magnitude.

(FALSE) The mechanical energy of the system decreases, because as the driver keeps the brakes on, the potential energy gravitational decreases when converted to thermal energy by friction, while kinetic energy remains constant.

(TRUE) Considering the system as a whole, energy is conserved, however, due to the friction of the brakes, part of the mechanical energy is transformed into thermal energy.

See too: Thermal energy

question 6

(UCB) A given athlete uses 25% of the kinetic energy obtained in running to perform a poleless high jump. If it reached a speed of 10 m/s, considering g = 10 m/s2, the height reached due to the conversion of kinetic energy into gravitational potential is as follows:

a) 1.12 m.
b) 1.25 m.
c) 2.5 m.
d) 3.75 m.
e) 5 m.

Correct alternative: b) 1.25 m.

Kinetic energy is equal to gravitational potential energy. If only 25% of the kinetic energy was used for a jump, then the quantities are related as follows:

25 percent sign. straight E with straight c subscript space equal to straight space E with straight p subscript space space 0 comma 25. diagonal numerator upwards straight line m. straight v squared over denominator 2 end of fraction equals diagonal space up straight line m. straight g. straight h space space numerator 0 comma 25 over denominator 2 end of fraction straight space v squared space equal to straight space g. straight h space 0 comma 125 straight space v squared space equal to straight space g. straight h space straight space h space equal to space numerator 0 comma 125 straight space v to the power of 2 space end of exponential over straight denominator g end of fraction

Replacing the values ​​of the statement in the formula, we have:

straight h space equal to space numerator 0 comma 125 space. space left parenthesis 10 straight space m divided by straight s right parenthesis squared space over denominator 10 straight space m divided by straight s ao square end of fraction straight space h space equal to numerator space 0 comma 125 space.100 straight space m squared divided by straight s squared over denominator 10 straight space m divided by straight s squared end of fraction straight h space equal to space numerator 12 comma 5 straight space m squared divided by straight s squared space over denominator 10 straight space m divided by straight s squared end of fraction straight h space equal to 1 comma 25 straight space m

Therefore, the height reached due to the conversion of kinetic energy into gravitational potential is 1.25 m.

See too: Potential energy

question 7

(UFRGS) For a given observer, two objects A and B, of equal masses, move with constant speeds of 20 km/h and 30 km/h, respectively. For the same observer, what is the reason?THE/ANDB between the kinetic energies of these objects?

a) 1/3.
b) 4/9.
c) 2/3.
d) 3/2.
e) 9/4.

Correct alternative: b) 4/9.

1st step: calculate the kinetic energy of object A.

straight E with straight A subscript space equal to numerator space left parenthesis straight m space. square space v ² right parenthesis space space over denominator 2 end of fraction straight E with straight A subscript space equal to numerator left parenthesis straight m space. space 20 ² right parenthesis space space over denominator 2 end of fraction straight E with straight A subscript space equal to numerator space left parenthesis straight m space. space 400 right parenthesis space over denominator 2 end of fraction straight E with straight A subscript space equal to space 200 space. straight space m

2nd step: calculate the kinetic energy of object B.

straight E with straight B subscript space equal to numerator space left parenthesis straight m space. straight space v ² right parenthesis over denominator 2 end of fraction straight E with straight B subscript space equal to numerator space left parenthesis straight m space. space 30 ² right parenthesis space space over denominator 2 end of fraction straight E with straight B subscript space equal to numerator space left parenthesis straight m space. space 900 right parenthesis over denominator 2 end of fraction straight E with straight B space subscript end of subscript equals space 450 space. straight space m

3rd step: calculate the ratio between the kinetic energies of objects A and B.

straight E with straight A subscript over straight E with straight B subscript space equal to numerator space 200 space. diagonal space up straight line m over denominator 450 space. diagonal space upwards straight line m end of fraction space straight space E with straight A subscript over straight E with straight B subscript space equal to space 200 over 450 space numerator divided by 50 over denominator divided by 50 end of fraction space straight E with straight A subscript over straight E with straight B subscript space equal to space 4 over 9

Therefore, reason ETHE/ANDB between the kinetic energies of objects A and B is 4/9.

See too: Kinetic energy

question 8

(PUC-RJ) Knowing that an 80 kg cybernetic runner, starting from rest, performs the 200 m test in 20 s maintaining a constant acceleration of a = 1.0 m/s², it can be said that the kinetic energy reached by the corridor at the end of 200 m, in joules, is:

a) 12000
b) 13000
c) 14000
d) 15000
e) 16000

Correct alternative: e) 16000.

1st step: determine the final speed.

As the runner starts from rest, its initial velocity (V0) has a value of zero.

straight V space equal to space straight V with 0 subscript space plus space at space straight space V space equal to space 0 space plus space 1 straight space m divided by straight s squared. space space 20 space straight space s straight V space equal to space 20 straight space m divided by straight s

2nd step: calculate the runner's kinetic energy.

straight E with straight c subscript space equal to numerator space left parenthesis straight m space. straight space v ² right parenthesis over denominator 2 end of fraction straight E with straight c subscript space equal to numerator space left parenthesis 80 space kg space. space left parenthesis 20 straight space m divided by straight space s right parenthesis ² right parenthesis space space over denominator 2 end of fraction straight E with straight c subscript space equal to space numerator left parenthesis 80 space kg space. space 400 straight space m squared divided by straight s squared right parenthesis over denominator 2 end of fraction straight E with straight c subscript space equal to numerator 32 space 000 over denominator 2 end of fraction space numerator kg space. straight space m squared over straight denominator s squared end of fraction straight E with straight c subscript space end of subscript equal to space 16 space 000 space numerator kg space. straight space m squared over straight denominator s squared end of fraction space equals space 16 space 000 straight space J

Thus, it can be said that the kinetic energy reached by the corridor at the end of the 200 m is 16 000 J.

question 9

(UNIFESP) A child weighing 40 kg travels in his parents' car, sitting in the back seat, fastened by the seat belt. At a given moment, the car reaches a speed of 72 km/h. Right now, the kinetic energy of this child is:

a) 3000 J
b) 5000 J
c) 6000 J
d) 8000 J
e) 9000 J

Correct alternative: d) 8000 J.

1st step: convert speed from km/h to m/s.

numerator 72 space km divided by straight h over denominator space 3 comma 6 end of fraction equal to space 20 straight space m divided by straight s

2nd step: calculate the child's kinetic energy.

Error converting from MathML to accessible text.

Therefore, the child's kinetic energy is 8000 J.

question 10

(PUC-RS) In a pole vault, an athlete reaches a speed of 11 m/s just before planting the pole in the ground to climb. Considering that the athlete can convert 80% of his kinetic energy into gravitational potential energy and that the gravity acceleration at the location is 10 m/s², the maximum height that its center of mass can reach is, in meters, about,

a) 6.2
b) 6.0
c) 5.6
d) 5.2
e) 4.8

Correct alternative: e) 4.8.

Kinetic energy is equal to gravitational potential energy. If 80% of the kinetic energy was used for a jump, then the quantities are related as follows:

80 percent sign. Ec space equal to space Ep space space 0 comma 8 space straight numerator m. straight v squared over denominator 2 end of fraction equals straight space m. straight g. straight h space space numerator 0 comma 8 over denominator 2 end of fraction straight space v squared space equal to straight space g. straight h space 0 comma 4 space. straight space v squared space equals straight space g. straight h space straight h space equal to space numerator 0 comma 4. straight v squared over straight denominator g end of fraction

Replacing the values ​​of the statement in the formula, we have:

straight h space equal to space numerator 0 comma 4 space. space left parenthesis 11 straight space m divided by straight s right parenthesis squared space space over denominator 10 straight space m divided by straight s squared end of fraction straight h space equal to space numerator 0 comma 4 space. 121 straight space m squared divided by straight s squared space over denominator 10 straight space m divided by straight s squared end of fraction straight h space equal to numerator 48 comma 4 straight space m squared divided by straight s squared space over denominator 10 straight space m divided by straight s squared end of fraction straight h space equal to space 4 comma 84 straight space m

Therefore, the maximum height that its center of mass can reach is approximately 4.8 m.

See too: Gravitational Potential Energy

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