PHYSICS A crate is pulled up a rough incline. The pulling force is parallel to t
ID: 1909236 • Letter: P
Question
PHYSICS A crate is pulled up a rough incline. The pulling force is parallel to the incline. The crate is pulled a distance of 6.77m. The acceleration of gravity is 9.8 m/s^2. Given: d = 6.77m theta = 18.2 m = 9.36 kg g = 9.8 coefficient of friction = .352 v = 1.45 m/s 1. What is the magnitude of the work done by the gravitational force? 2. Is the work done by the gravitational field zero, positive, or negative? 3. How much work is done by the 127 N force? 4. What is the change in kinetic energy of the crate? 5. What is the speed of the crate after it is pulled 6.77 m?Explanation / Answer
W = w1 + w2
W = mgsin*(-6.77) + mgcos (0)
= (9.36)(9.8)sin(18.2)(-6.77) + 0
= - 193.9 J
The work done by gravitational force = -193.9 J
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W(F) = F.S
= (127 ) (6.77)
= 859.79 J
The work done by the Force = 859.79J
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Initial kinetic energy = [m(u)^2] / 2
K.E (i) = [9.36(1.45)^2 ] / 2
K.E (i) = 9.83 J
Rf = F - mgsin - mgcos
= 127 - (9.36)(9.8)sin(18.2) - 30.67
= 67.68 N
acc = Rf / mass
= 67.68 / 9.36
= 7.23 N
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