How can I solve for acceleration in this kind of questions?

Question

A 34.7 kg box is towed up a hill inclined at 15.3 degrees with respect to the horizontal. The tow rope is parallel to the incline and exerts a force of 104 N on the box. Assume that the box starts from rest at the bottom of the hill, and disregard friction.
The acceleration of gravity is 9.81 m/s^2.
How fast is the bos going after moving 45.1m up the hill? Answer in units of m/s^2.
Please explain it.
Thank you.

Answer 1

draw a FBD and you'll see Fweight(x) of the box and Tension are the Fnet.

because the velocity changes, we have a acceleration

a = Fnet/m

a = ( Tension - Fweight(x) )/ m

break down Fweight(x)
Fweight(x) = sin(15.3)*mg

a = (Tension - sin(15.3)*mg) / m

a= ( 104 - sin(15.3)*(34.7*9.81) ) / 34.7

a = 0.409m/s^2

use acceleration we just found to find the velocity
Vf^2 = 2ad + Vi^2
Vf^2 = 2(0.409)(45.1) + 0^2
Vf^2 = 36.892
Vf = 6.07m/s

hope this helps

Answer 2

Errmmm..... Something is not quite right. m/s² are the units for acceleration, not speed (which is m/s)

But.... If the box is on an incline of 15.3 degrees, then the gravitational component is going to be
F1 = 34.7*9.8*sin(15.3) = 89.73 N acting 'downhill' and parallel to the inclined polane. There is a force of 104 N acting 'uphill' and parallel to the inclined plane so the total force acting on the box is
104 - 89.73 = 14.27 N going 'uphill' so the box is accelerating at
14.27/34.7 = .411 m/s² After 45.1 meters, it's speed will be
√(2*.411*45.1) = 6.089 m/s

HTH ☺


Doug