A stop block, s, prevents a crate from sliding down a theta = 27.0 degree inclin

Question

A stop block, s, prevents a crate from sliding down a theta = 27.0 degree incline. (Figure 1)A tensile force F = (F_0t)N acts on the crate parallel to the incline, where F_0 = 350 N/s. If the coefficients of static and kinetic friction between the crate and incline are meu_s = 0.285 and meu_k = 0.185, respectively, and the crate has a mass of 56.4 kg, how long will it take until the crate reaches a velocity of 3.16 m/s as it moves up the incline? Express your answer numerically in seconds to three significant figures. t= s

Explanation / Answer

Here,

theta = 27 degree

F = 350 * t

Using second law of motion

static friction , Fs = us * mg * cos(theta)

Fs = 0.285 * 56.4 * 9.8 * cos(27)

Fs = 140.4 N

kinetic friction, Fk = uk * mg * cos(theta)

Fk = 0.185 * 56.4 * 9.8 * cos(27)

Fk = 91.1 N

the cart will start moving when

Fs + mg * sin(theta) = F

350 * t = 140.4 + 56.4 * sin(27) * 9.8

solving for t

t = 1.12 s

Now, for the acceleration is a

using second law of motion

56.4 * a = F - Fk - mg * sin(theta)

56.4 * a = 350 * (t - 1.12) - 91.1 - 56.4 * 9.8 * sin(27)

a = 6.20 * t - 13.01

v = integartion(6.2 * t - 13.01) from t = 1.12 s to t

v = (3.1 * t^2 - 13.01 * t)

3.16 = (3.1 * t^2 - 13.0* t - (3.1 * 1.12^2 - 13.01 * 1.12))

sovling for t

t = 3.5 s

the time taken for the crate to reach 3.16 m/s is 3.5 s

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