Blocky is sliding down a hill. His mass is 15 kg, and the hill has an angle of 35 degrees. If the hill is 80 m long, find the following.

Question

a) Blocky’s velocity at the bottom of the hill.

b) The time it takes for blocky to reach the bottom of the hill.

This is Physics 1

Please show work and explain

Answer 1

This problem should be split into two parts, what's happening in the x direction and what's happening in the y direction.

First, draw a sketch (including the coordinate plane we are going to use):

Then draw a free body diagram (since friction is not mentioned, we'll ignore friction). A free body diagram shows the object in question as a dot and then all the forces acting on it - in this case, the forces are gravity and the normal force (the force the ramp is putting on the block that keeps the block from falling through the ramp). However, as I mentioned, we will divide the problem into two parts, the x direction and the y direction, so, since gravity, F_g, is not aligned to either the x or the y direction, we will split it into it's vector components that do align with the x and y directions:

Using trigonometric ratios, we can find that sin(35°) = F_g-x/F_g or F_g-x = F_gsin(35°)

Since we know that the force of gravity equals the mass times the acceleration of gravity (g) and that g = 9.81 m/s², Fg = (15)(9.81) = 147.15 N

Then F_g-x = (147.15)(sin(35°)) = 84.4 N

Since that's the force in the direction of the travel, that's the force that will be pulling Blocky down the hill, To find his acceleration, we use F = ma or a = F/m = 84.4/15 = 5.627 m/s²

Now, we use the kinematic equations to find his velocity using this acceleration. v_f² = v_i² + 2ax. We will assume he starts from rest (v_i = 0) so v_f² = 0² + 2(5.627)(80) or .v_f² = 900.29 and, taking the square root, we get v_f = 30.005 m/s or rounding to 2 significant figures, v_f = 30 m/s

To get the time, we use a different kinematic equation: v_f = v_i + at

So 30.005 = 0 + 5.627t and, solving for t, we get t = 5.332 s, again, rounding to 2 significant figures, we get t = 5.3 s