A block is sent up a frictionless ramp along which an x-axis extends
upward. The figure below gives the kinetic energy of the block as a function of position x. If the block’s initial speed is 4.00 m/s, what is the normal force on the block? (initial KE is 40J)
initially i though this:
KE=1/2mv^2
so 40=1/2m(4^2)
so m=5kg Fn=Mgcos(theta)
so Fn=5*9.8*cos(theta).... but what is theta?
1 answer
However, the normal force on the block is not dependent on the block's initial speed. The normal force is equal to the weight of the block multiplied by the cosine of the angle of the ramp.