A block is released from rest at the top of a frictionless ramp (inclined plane)

Question

A block is released from rest at the top of a frictionless ramp (inclined plane) that makes an angle of 25° with respect to horizontal. At the bottom of the ramp, after having slid a distance D1, the speed of the block is observed to be 12 m/s. What is the length of the ramp D1?   

Part A:

The block now slides onto the horizontal floor with the same initial speed of 12m/s. The coefficient of kinetic friction between the block and the floor is k =0.3. From the moment it reaches the floor, how much time does it take the block to stop?

Part B

. If the mass of the block in the above problem were doubled, the time required for the block to stop would

(a)Increase                                      (b) Decrease                                 (c) Stay the same

Choose one.

Explanation / Answer

here,

angle of ramp, A = 25 degrees
Velocity at bottom, v = 12 m/s

From Conservation fo Energy
Loss in potential Energy = Gain in Kinetic Energy
mg*h = 0.5 * m*v^2

also from trigonometry,
Sin25*d = h

rewriting energy eqn in terms of distance travelled, d

d = 0.5*v^2/Sin25*g
d = (0.5*12^2)/(Sin25*9.8)
d = 17.384 m

Part B:
uk = 0.3

From newton law of motion, Fnet = 0

F - Frictional Force = 0
ma = uk*mg

solving for acceleation of block, a

a = uk*g --------------------(1)
a = 0.3*9.8
a = 2.94 m/s^2

From First eqn of motion,
v = at

solving time taken by block to stop,
t = v/a
t = 12/2.94
t = 4.082 s -------------------(2)

Part C:
as seen in eqn 1 and 2 system is independent of mass so time will be same , but if acceleration due to gravity is changed then there will change in time will noticed

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