Question

1. A cart of mass 2.0 kg is released off the floor from a distance of 3.5 m. What is the change in the potential energy of the ball? Show the calculations

. 2. Later the same cart is released from rest, and it falls on the floor. What is the kinetic energy of the cart just before it hits the floor? What is its velocity before hitting? Show your calculations.

3. a) Suppose, an object is released in the air. The object attains the highest point, and falls down back again and is caught at the height it was released. When will the potential energy of the object have its maximum and minimum value? Explain.

b) In case of kinetic energy, when will the kinetic energy be maximum?

Answer 1

1)Given,

m = 2 kg ; h = 3.5 m

initial PE = 0

final PE = m g h = 2 x 9.8 x 3.5 = 68.6 J

change in PE = final - intial

change in PE = 68.6 - 0 = 68.6 J

delta-PE = 68.6 J

2)from conservation of energy

KE = PE = 68.6 J

KE = 1/2 m v^2 = 68.6

v = sqrt (2 x 68.6/2) = 8.28 m/s

Hence, v = 8.28 m/s

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3)(a)The potential energy of the object will be maximum at the highest point. As we know that, PE = m g h, so at highest point PE will be max and will be zero at ground level (lowest point).

b)The kinetic energy will be maximumat the lowest point as the speed will be maximum there. KE = 1/2 m v^2. At highest point the speed is zero and hence KE is lowest there.