Physics, 07.03.2020 05:59 sportie59

A crate of mass 10.0 kg is pulled up a rough incline with an initial speed of 1.50 m/s. The pulling force is 100 N parallel to tthe incline, which makes an angle of 20.0 degrees with the horizontal. The coefficient of kinetic friction is 0.400, and the crate is pulled 5.00 mm.

(a) How much work is done by the gravitational force on the crate? (b) Determine the increase in internal energy of the crate-incline system owing to friction. (c) How much work is done by the 100- N force on the crate? (d) What is the change in kinetic energy of the crate? (e) What is the speed of the crate after being pulled 5.00 m?

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Answer from: itsdeevv

(a) work done by the gravitational force on the crate is -167.59J

(b) the increase in internal energy of the crate-incline system owing to friction is 184.2 J

(d) the change in kinetic energy of the crate is 118.06 J

(e) the speed of the crate after being pulled 5.0 m is 5.09 m/s

Explanation:

Given;

mass of the crate, m = 10.0 kg

initial speed of the crate, u = 1.50 m/s

parallel force on the crate, F = 100N

angle of inclination of the force, θ = 20°

coefficient of friction, μ = 0.4

distance moved by the crate, h = 5.0m

Part (a) work done by the gravitational force on the crate:

This work is due to vertical displacement of the crate = h*sinθ,

W = -(mgh*sinθ)= -(10*9.8*5.0*sin(20))= -167.59 J

This work is negative because it is acting against the crate.

Part (b) the increase in internal energy of the crate-incline system owing to friction.

Magnitude of frictional force = μ*N

where;

N is the normal reaction = mgcosθ

Magnitude of frictional force = μ*mgcosθ

= 0.4*10*9.8*cos(20) = 36.84 N

Thus, increase in internal energy of the the crate-incline system owing to friction = Magnitude of frictional force X h

= 36.84N x 5m

= 184.2 J

Part (c) work done by the 100- N force on the crate

W = Fd*cosθ = 100*5*cos(20) = 469.85 J

Part (d) the change in kinetic energy of the crate:

ΔK.E = W.force - W.gravity - W.friction

ΔK.E = 469.85 J - 167.59 - 184.2 J = 118.06 J

Part (e) the speed of the crate after being pulled 5.0 m

Δ.KE $= K.E_f-K.E_i\\\\$

$\delta K.E= \frac{1}{2}mv_f^2 -\frac{1}{2}mv_i^2\\\\\frac{1}{2}mv_f^2 = \delta K.E + \frac{1}{2}mv_i^2\\\\v_f = \sqrt{\frac{2\delta K.E}{m} +v_i^2}\\\\v_f = \sqrt{\frac{2*118.06}{10} +(1.5)^2} =5.09 m/s$

v = 5.09 m/s

A crate of mass 10.0 kg is pulled up a rough incline with an initial speed of 1.50 m/s. The pulling

Answer from: Quest

hi there!

out of the two, (b) precision would be more reasonable. however, both a and b could be pretty reasonable.

-awesomerepublic : )

Answer from: Quest

Idont understand the question

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