Question

We want to slide a 12 kg crate up a 2.5 m long ramp inclined at 30o. A worker, ignoring friction, calculates that he can do this by giving it an initial speed of 5.0 m/s at the bottom and letting it go. But friction is not negligible; the crateslides only 1.6 m up the ramp, stops, and slides back down.

Find the magnitude of the friction force acting on the crate, assuming that it is constant.

Answer 1

Answer:

33.72N

Explanation:

To find the magnitude of the friction force you take into account all forces over the object:

$-mgsin\theta - F_f+F=ma$    (1)

where Mgsin0 is the gravitational force, Ff is the friction force F is the force of the worker, m is the mass and a is the acceleration of the object while it is going up the ramp.

it is necessary to find the acceleration of the crate:

$v_f^2=v_o^2-2ad\\\\a\frac{v_f^2-v_o^2}{2d}=\frac{-(5.0m/s)^2}{2(1.6)m}=7.81\frac{m}{s^2}$

with this value and the values for m, g, the angle and d you obtain, by using (1)  you obtain:

$-(12)(9.8)sin30\°-F_f+F=(12)(7.81)\\\\-F_f+F=152.52$(3)

For the calculation of F:

If the surface would be frictionless you have:

$-mgsin30\°+F=ma\\\\a=\frac{v_o^2}{2(2.5)m}=5\frac{m}{s^2}$

with these values you can calculate the force F:

$F=ma+mgsin30\°=(12kg)(5m/s^2+9.8m/s^2sin30\°)=118.8N$

BY replacing this values in the expression (3) you get:

$F_f=152.52-118.8N=33.72N$

hence, the magnitude of the friction force is 33.72N

Answer 2

The magnitude of the friction force acting on the crate is 33.4 N.

The given parameters;

• mass of the crate, m = 12 kg
• length of the ramp, d = 2.5 m
• angle of the ramp, θ = 30⁰
• initial speed of the crate, u = 5 m/s
• length of the frictionless ramp, d = 1.6 m

The acceleration of the crate when limited by friction on the ramp is calculated as;

$v^2 = v_0^2 + 2(-a)d\\\\v^2 = v_0^2 - 2ad\\\\0 = v_0^2 - 2ad\\\\2ad = v_0^2 \\\\a = \frac{v_0^2}{2d} \\\\a = \frac{5^2}{2(1.6)} \\\\a= 7.81 \ m/s^2$

The net force on the crate is calculated as;

$\Sigma F_{net} = ma\\\\-mg sin\ \theta \ - F_f \ + F = ma\\\\-F_f + F = ma + mgsin\ \theta \\\\-F_f + F = 12(7.81) \ + \ (12 \times 9.8)sin(30)\\\\-F_f + F = 152.52 \ N$

The acceleration of the crate on frictionless ramp is calculated as;

$a = \frac{v_0^2}{2d} \\\\a = \frac{(5)^2}{2(2.5)} = 5 \ m/s^2$

The net force on the crate is calculated as;

$-mg sin\ \theta + F = ma\\\\F =ma + mg sin \ \theta \\\\F =(12)(5) \ + \ (12)(9.8)sin \ (30)\\\\F = 118.8 \ N$

The magnitude of the friction force acting on the crate is calculated as;

$-F_f + F = 152.52\\\\F_f = F - 152.52\\\\F_f = 118.8 \ - \ 152.52\\\\F_f = -33.4 \ N\\\\|F_f| = 33.4 \ N$

Thus, the magnitude of the friction force acting on the crate is 33.4 N.

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