Answer:
987 joules, 3.01s
Explanation:
(A)
from the attached diagram
net force, Fnet, pulling the crate up the ramp is given by
Fnet = FcosФ - WsinФ - Fr
where FcosФ is the component of horizontal force 290N resolved parallel to the plane
WsinФ = mgsinФ = component of the weight of the crate resolved parallel to the plane
Fr = constant opposing frictional force
Fnet = 290cos34⁰ - 20 × 9.8 × sin34° - 65
Fnet = 240.421 - 109.602 - 65
Fnet = 65.82N
Work done on the crate up the ramp, W, is given by
W = Fnet × d (distance up the plane)
W = 65.819 × 15
W = 987.285 joules
W = 987 joules (to 3 significant Figures)
(B)
to calculate the time of travel up the ramp
we use the equation of motion
![s = ut + \frac{1}{2}at^{2}](https://tex.z-dn.net/?f=s%20%3D%20ut%20%2B%20%5Cfrac%7B1%7D%7B2%7Dat%5E%7B2%7D)
where s = distance up the plane, 15m
u = Initial velocity of the crate, which is 0 for a body that is initially at rest
a = acceleration up the plane, given by
![a = \frac{Fnet}{m}](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7BFnet%7D%7Bm%7D)
where m = mass of the crate, 20 kg
now, ![a = \frac{65.819}{20} \\a = 3.291\frac{m^{2} }{s}](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7B65.819%7D%7B20%7D%20%5C%5Ca%20%3D%203.291%5Cfrac%7Bm%5E%7B2%7D%20%7D%7Bs%7D)
from, ![s = ut + \frac{1}{2}at^{2}](https://tex.z-dn.net/?f=s%20%3D%20ut%20%2B%20%5Cfrac%7B1%7D%7B2%7Dat%5E%7B2%7D)
![15 = 0*t + \frac{1}{2}* 3.291 * t^{2}](https://tex.z-dn.net/?f=15%20%3D%200%2At%20%2B%20%5Cfrac%7B1%7D%7B2%7D%2A%203.291%20%2A%20t%5E%7B2%7D)
15 = 0 + 1.645![t^{2}](https://tex.z-dn.net/?f=t%5E%7B2%7D)
15 = 1.645![t^{2}](https://tex.z-dn.net/?f=t%5E%7B2%7D)
![t = \sqrt{\frac{15}{1.645} }](https://tex.z-dn.net/?f=t%20%3D%20%5Csqrt%7B%5Cfrac%7B15%7D%7B1.645%7D%20%7D)
![t = 3.019](https://tex.z-dn.net/?f=t%20%3D%203.019)
t = 3.01s (to 3 sig fig)