A girl having mass of 35 KG sits on a trolley of mass 5 KG. The trolley is given an initial velocity of 4 m/s by applying a force. The trolley comes to rest after travelling a distance of 16 m.
(i) How much work is done on the trolley?
(ii) How much work is done by the girl?
work done by girl is zero,because she is sitting on the trolley and force is perpendicular to displacement.
but in book it is given 280J.
How is it?
Mass of girl = 35 kg
Mass of trolley = 5 kg
Total mass, m = (35 + 5) = 40 kg
u = 4 m/s
v = 0
s = 16 m
By equation,
v 2 = u 2 + 2as
0 = (4)2 + 2a (16)
32a = –16
a = –0.5 m/s2
Force exerted on trolley,
F = ma
= 40 × 0.5 = 20 N
Work done on trolley W = FS = 20 N x 16 m
= 320 J
Work done by the girl, W = FS
= mass of girl × retardation × S
= 35 × 0.5 × 16
= 280 J
in each case, the object traveling has initial kinetic energy, and work is done by each on friction that equals this KE.
To find the answers to these questions, we need to understand the concept of work done. Work is defined as the product of force applied and the displacement of an object in the direction of the force. Mathematically, work (W) is given by the equation:
W = F * d * cos(θ)
Where:
- F is the force applied
- d is the displacement
- θ is the angle between the force and displacement vectors.
In this scenario:
(i) To find the work done on the trolley, we need to calculate the net force acting on the trolley. The net force is equal to the product of the mass of the system (trolley + girl) and the acceleration of the system. Since the trolley comes to rest, the final velocity is 0. Therefore, we can use the equation:
F_net = m * a
The acceleration, in this case, can be calculated using the equation of motion:
v^2 = u^2 + 2 * a * d
Where:
- v is the final velocity (0 m/s)
- u is the initial velocity (4 m/s)
- a is the acceleration
- d is the displacement (16 m)
Rearranging the equation, we can find the acceleration:
0 = 4^2 + 2 * a * 16
16a = -16
a = -1 m/s^2
Now, we can calculate the net force:
F_net = (m_trolley + m_girl) * a
F_net = (5 + 35) * (-1)
F_net = -40 N
Since the displacement and force are acting in the same direction, the angle θ between them is 0 degrees. Substituting the values into the equation, we get:
W_trolley = F * d * cos(θ)
W_trolley = -40 * 16 * cos(0)
W_trolley = -40 * 16
W_trolley = -640 J
Therefore, the work done on the trolley is -640 Joules.
(ii) To find the work done by the girl, we need to consider only the force she applies to the trolley. In this case, the force applied by the girl (F_girl) is equal to her mass multiplied by the acceleration:
F_girl = m_girl * a
F_girl = 35 * (-1)
F_girl = -35 N
Again, since the displacement and force are acting in the same direction, the angle θ between them is 0 degrees. Substituting the values into the equation, we get:
W_girl = F_girl * d * cos(θ)
W_girl = -35 * 16 * cos(0)
W_girl = -35 * 16
W_girl = -560 J
Therefore, the work done by the girl is -560 Joules.