A block of ice is sliding down a ramp of slope 40 degrees to the horizontal. What is the rate of acceleration of the block?

Question

A block of ice is sliding down a ramp of slope 40 degrees to the horizontal. What is the rate of acceleration of the block?

Answer 1

Because they tell you that this is a block of ice, you are likely supposed to assume that block's motion down the plane is frictionless. The component of gravitational force in parallel to the ramp is the only unbalanced force acting on the block, and its magnitude is mg*sin(θ). Applying Newton's Second Law, F=ma⇔mg*sin(θ)=ma⇔a=g*sin(θ).

Plug in the given value for theta and acceleration due to gravity.

Answer 2

If sum of the forces F = ma, where m = the mass, & a = the acceleration of the block, b = angle = 40 degrees then

(mg)sin(b) = ma or a = (g)sin(b) = (32.2)sin(40) = 20.7 ft/sec²

where g = the acceleration due to gravity (32.2 ft/sec²)