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Work and Energy | Work and Energy - Exercise

Question 23

A girl having a mass of 35 kg sits on a trolley of mass 5 kg. The trolley is given an initial velocity of 4 m s–1 by applying a force. The trolley comes to rest after traversing a distance of 16 m. (a) How much work is done on the trolley? (b) How much work is done by the girl?

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  • Solution

  • Transcript

Initial velocity of the trolley,u = 4 m/s

Final velocity of the trolley v = 0

Mass of the trolley m = 5 kg

Distance covered by the trolley before coming to rest,s = 16 m

From Equation 2 as = v2-u2,

a = v2-u22 S

= 0-(4)22×16

= 0.5 m/s2

Force (frictional) acting on the trolley = ma

= 40 (- 0.5)

= – 20 N

Work done on the trolley = Fs = (20 N) (16 m)

= 320 J

(b) Since the girl does not move w.r.t. the trolley (as she is sitting on it), work done by the girl = 0.

"hello everyone welcome to lido learning i am gurpreet your science tutor today's question is a girl having a mass of 35 kg sits on a trolley of mass 5 kg the trolley is given an initial velocity of 4 meter per second by applying a force just trolley comes to rest after traversing a distance of 16 meters how much work is done on the trolley and how much work is done by the girl so here in this question we are going to find out that what is the amount of work done on the trolley and what is the work done by the girl so let's start our answer here in the question initial velocity of trolley is given that is u is equals to 4 meter per second and final velocity of the trolley is 0 because it is coming to rest and mass of the trolley is given as 5 kgs and distance covered that is s will be equals to 16 meter so according to the trolley we'll calculate the acceleration by the third equation of motion that is 2 a s is equal to v square minus q square from this formula will calculate the value of acceleration for trolley so a will be equals to v square minus u square by 2s fine now here v square that is 0 minus u square that is 4 square fine now upon 2 into s that is 2 into 60 so you will get the acceleration as 0.5 meter per second square okay after calculating the acceleration we will calculate the value of fourth for the trolley the force by which it was able to stop and come to rest that was a frictional force and this will be equals to the mass into acceleration so here mass is 5 kg 5 kg plus 35 kg of the girl that is 40 and acceleration we have calculated as minus 0.5 meter per second so you will get the frictional force as minus 20 newtons now work done by the trolley for the a part will calculate work done the troll this will be equal to force into distance and distance here is 16 meter force we have calculated minus 20 so we'll put without the negative sign that is 20 into 60 will get the work done as 320 joules fine now for the b part what will be the work done by the girl here the girl does not move the girl is not moving only the trolley is moving girl is just sitting on the trolley so work done by the girl will be zero for more such videos please subscribe leader learning and for any doubts drop a comment thank you"

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Question 23

A girl having a mass of 35 kg sits on a trolley of mass 5 kg. The trolley is given an initial velocity of 4 m s–1 by applying a force. The trolley comes to rest after traversing a distance of 16 m. (a) How much work is done on the trolley? (b) How much work is done by the girl?

  • Solution

  • Transcript

Initial velocity of the trolley,u = 4 m/s

Final velocity of the trolley v = 0

Mass of the trolley m = 5 kg

Distance covered by the trolley before coming to rest,s = 16 m

From Equation 2 as = v2-u2,

a = v2-u22 S

= 0-(4)22×16

= 0.5 m/s2

Force (frictional) acting on the trolley = ma

= 40 (- 0.5)

= – 20 N

Work done on the trolley = Fs = (20 N) (16 m)

= 320 J

(b) Since the girl does not move w.r.t. the trolley (as she is sitting on it), work done by the girl = 0.

"hello everyone welcome to lido learning i am gurpreet your science tutor today's question is a girl having a mass of 35 kg sits on a trolley of mass 5 kg the trolley is given an initial velocity of 4 meter per second by applying a force just trolley comes to rest after traversing a distance of 16 meters how much work is done on the trolley and how much work is done by the girl so here in this question we are going to find out that what is the amount of work done on the trolley and what is the work done by the girl so let's start our answer here in the question initial velocity of trolley is given that is u is equals to 4 meter per second and final velocity of the trolley is 0 because it is coming to rest and mass of the trolley is given as 5 kgs and distance covered that is s will be equals to 16 meter so according to the trolley we'll calculate the acceleration by the third equation of motion that is 2 a s is equal to v square minus q square from this formula will calculate the value of acceleration for trolley so a will be equals to v square minus u square by 2s fine now here v square that is 0 minus u square that is 4 square fine now upon 2 into s that is 2 into 60 so you will get the acceleration as 0.5 meter per second square okay after calculating the acceleration we will calculate the value of fourth for the trolley the force by which it was able to stop and come to rest that was a frictional force and this will be equals to the mass into acceleration so here mass is 5 kg 5 kg plus 35 kg of the girl that is 40 and acceleration we have calculated as minus 0.5 meter per second so you will get the frictional force as minus 20 newtons now work done by the trolley for the a part will calculate work done the troll this will be equal to force into distance and distance here is 16 meter force we have calculated minus 20 so we'll put without the negative sign that is 20 into 60 will get the work done as 320 joules fine now for the b part what will be the work done by the girl here the girl does not move the girl is not moving only the trolley is moving girl is just sitting on the trolley so work done by the girl will be zero for more such videos please subscribe leader learning and for any doubts drop a comment thank you"

Our top 5% students will be awarded a special scholarship to Lido.

subject-cta
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