Question

crate is moved across the floor by pulling a
rope tied to it. The force on the crate is of
magnitude 450 N and is directed at an angle
of 60° to the horizontal. The force of friction
on the crate is 125 N. The mass of the crate is
300 kg. Draw the free body diagram for the
system and calculate the acceleration of the
crate. Calculate the work done by each of the
forces in 'displacing the crate by 5.0 m. Which
of these forces is a "no-work" force ?

Answer 1

Answer on Question #73705, Physics / Mechanics | Relativity

Question. Crate is moved across the floor by pulling a rope tied to it. The force on the crate is of magnitude $450N$ and is directed at an angle of $60{}^{\circ}$ to the horizontal. The force of friction on the crate is $125N$ . The mass of the crate is $300kg$ . Draw the free body diagram for the system and calculate the acceleration of the crate. Calculate the work done by each of the forces in displacing the crate by $5.0m$ . Which of these forces is a "no-work" force?

Solution.

According to the second Newton's low

$$\vec {F} = m \vec {a}.$$

So

$$\sum F _ {x} = m a$$ $$F \cos \alpha - F _ {f} = m a \rightarrow a = \frac {F \cos \alpha - F _ {f}}{m} = \frac {4 5 0 \cdot \cos 6 0 {}^ {\circ} - 1 2 5}{3 0 0} = 0. 3 3 \frac {m}{s ^ {2}}.$$ $$\sum F _ {y} = 0.$$ $$N + F \sin \alpha - F _ {g} = 0 \rightarrow N + 4 5 0 \cdot \sin 6 0 {}^ {\circ} - 3 0 0 \cdot 9. 8 = 0 \rightarrow N = 2 9 4 0 - 3 9 0 = 2 5 5 0 N.$$ $$A = \left(\vec {F} \cdot \vec {r}\right) = F \cdot r \cdot \cos \alpha = 4 5 0 \cdot 5 \cdot \cos 6 0 {}^ {\circ} = 1 1 2 5 J.$$ $$A _ {f} = \left(\vec {F} _ {f} \cdot \vec {r}\right) = F _ {f} \cdot r \cdot \cos 1 8 0 {}^ {\circ} = 1 2 5 \cdot 5 \cdot (- 1) = - 6 2 5 J.$$

$\vec{N}$ and $\vec{F}_g$ are "no-work" forces.

Answer. $a = 0.33\frac{m}{s^2};A = 1125J;A_f = -625J.$

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