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**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?**

Answer:

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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