A 2-kg block is released from rest at the top of a 20-mlong frictionless ramp that is 4 m high. At the same time, an identical block is released next to the ramp so that it drops straight down the same 4 m. What are the values for each of the following for the blocks just before they reach ground level. Required:
a. Gravitational potential energy Block a J Block b J
b. Kinetic energy Block a J Block b
c. Speed Block a J Block b J
d. Momentum Block a J Block b J

A.) 78.4 J for both

B.) 78.4 J for both

C.) 8.85 m/s for both

D.) 17.7 kgm/s

Explanation:

Given information:

Mass m = 2 kg

Distance d = 20 m

High h = 4 m

A.) Gravitational potential energy can be calculated by using the formula

P.E = mgh

P.E = 2 × 9.8 × 4

P.E = 78.4 J

Since the two objects are identical, the gravitational potential energy of the block for both a and b will be 78.4 J

B.) According to conservative energy,

Maximum P.E = Maximum K.E.

Therefore, the kinetic energy of the two blocks will be 78.4 J

C.) Since K.E = 1/2mv^2 = mgh

V = √(2gh)

Solve for velocity V by substituting g and h into the formula

V = √(2 × 9.8 × 4)

V = √78.4

V = 8.85 m/s

The velocities of both block will be 8.85 m/s

D.) Momentum is the product of mass and velocity. That is,

Momentum = MV

Substitute for m and V into the formula

Momentum = 2 × 8.85 = 17.7 kgm/s

Both block will have the same value since the ramp Is frictionless.