Question

A mass of 3.8 kg is originally moving at 8 m/s at the top of a frictionless incline which has a length of 6.2 meters and an inclination angle of 56 degrees. It slides down the incline and over a horizontal surface in which a portion of it has friction. The coefficient of kinetic friction is 0.37 and the portion of the surface which has friction is 8 meters. At the end of the horizontal surface is a spring. The mass compresses the spring 0.65 meters before it is stopped. What is the amount of force produced by the spring when the mass is stopped in Newtons?

Answer 1

Gravitational acceleration = g = 9.81 m/s²

Mass of the object = m = 3.8 kg

Speed of the object at the top of the incline = V₁ = 8 m/s

Length of the incline = L = 6.2 m

Angle of incline = = 56^o

Height of the incline= H

H = LSin

Speed of the object at the bottom of the incline = V₂

By conservation of energy the potential and kinetic energy of the object at the top of the incline is equal to the kinetic energy of the object at the bottom of the incline.

mV₁²/2 + mgH = mV₂²/2

V₁²/2 + gLSin = V₂²/2

(8)²/2 + (9.81)(6.2)Sin(56) = V₂²/2

V₂ = 12.84 m/s

Coefficient of kinetic friction of the rough floor = = 0.37

Friction force on the object = f = mg

Length of the rough floor = D = 8 m

Speed of the object after crossing the rough floor = V₃

By conservation of energy the kinetic energy of the object before the rough floor is equal to the kinetic energy of the object after passing the rough floor plus the work done against friction.

mV₂²/2 = mV₃²/2 + fD

mV₂²/2 = mV₃²/2 + mgD

V₂²/2 = V₃²/2 + gD

(12.84)²/2 = V₃²/2 + (0.37)(9.81)(8)

V₃ = 10.334 m/s

Spring constant = k

Compression of the spring when the object comes to a stop = X = 0.65 m

By conservation of energy the kinetic energy of the object before coming in contact with the spring is converted into the potential energy of the spring as the object comes to a stop.

mV₃²/2 = kX²/2

mV₃² = kX²

(3.8)(10.334)² = k(0.65)²

k = 960.492 N/m

Force applied by the spring when the mass is stopped = F

F = kX

F = (960.492)(0.65)

F = 624.32 N

Force applied by the spring when the mass is stopped = 624.32 N