Question

5. A box is released on a smooth incline, which makes an angle of 37o with respect to the horizontal.

a) What is the boxs acceleration?

b) The box starts from rest and slides a distance (along the incline) of 75 cm to the bottom of the ramp. What is the speed of the box at the bottom?

c) The mass of the box is 100 g. When it gets to the bottom it compresses a spring with a force constant 1000 N/m. What is the maximum compression of the spring?

Answer 1

(a) = 37 deg

Acceleration of the box, a = g*sin = 9.8*sin37 = 5.9 m/s^2

(b) distance traveled by the box along the incline, d = 75 cm = 0.75 m

So, velocity of the box at the bottom, v = sqrt[2*a*d]

= sqrt[2*5.9*0.75] = 2.97 m/s (Answer)

(c) Mass of the box, m = 100 g = 0.10 kg

Kinetic energy of the box at the bottom, KE = (1/2)*m*v^2

this energy will be stored in the spring as spring energy.

Spring energy, SE = (1/2)*k*x^2

apply conservation of energy -

KE = SE

=> (1/2)*m*v^2 = (1/2)*k*x^2

=> x^2 = [m*v^2] / k = [0.10*2.97^2] / 1000

=> x = 0.0297 m (Answer)