Fap = 131N = Applied force.
Fc=mg=14kg * 9.8N/kg=137.2N @12.3deg.
Fp = 137.2sin12.3 = 29.2N = Force parallel to the plane.
Fv = 137.2cos12.3 = 134.1N = Force
perpendicular to the plane.
Ff = u*Fv = 0.26 * 134.1 = 34.9N = Force of friction.
Fn = Fap - Fp - Ff,
Fn = 131 - 29.2 - 34.9 = 66.9N = Net force acting on crate.
a. W=Fc*h = 137.2 * 8.4sin12.3 = 246J.
b. W = Ff*d = -34.9 * 8.4 = 293J.
c. W = Fap * d = 131 * 8.4 = 1100J.
I will have to do some research on d and e. But I hope this will help.
Can anyone explain this to me, it was on my exam and i don't get it ;
A 14.0 kg crate is pulled up a rough incline with an initial speed of 1.6 m/s. The pulling force is 131.0 N parallel to the incline, which
makes an angle of 12.3 ◦ with the horizontal. The coefficient of kinetic friction is 0.26 and the crate is pulled a distance of 8.4 m. The acceleration of gravity is 9.81 m/s
a) Find the work done by Earth’s gravity
on the crate.
Answer in units of J.
b) Find the work done by the force of friction
on the crate.
Answer in units of J
c) Find the work done by the puller on the
crate.
Answer in units of J.
d) Find the change in kinetic energy of the
crate.
Answer in units of J
e) Find the speed of the crate after it is pulled 8.4 m.
Answer in units of m/s.
1 answer
1 answer