Friction pulls directly against the direction of motion (at 180) of a sled, and does -55.2 J of work while the sled moves 8.98 m

Question

3 years ago

12

. What is the magnitude (+) of the friction force?(unit=N)

1 answer:

zvonat [6] · Answer 1 · 2022-05-16T06:56:06+03:00

The magnitude of the force of friction is 6.1 N

Explanation:

The work done by a force when moving an object is given by the equation:

$W=Fd cos \theta$

where :

F is the magnitude of the force

d is the displacement

$\theta$ is the angle between the direction of the force and of the displacement

In this problem, we have:

W = -55.2 J (work done by the force of friction)

d = 8.98 m (displacement of the sled)

$\theta=180^{\circ}$ (because the force of friction acts opposite to the direction of motion)

By solving the equation for F, we find the magnitude of the force of friction on the sled:

$F=\frac{W}{d cos \theta}=\frac{-55.2}{(8.98)(cos 180^{\circ})}=6.1 N$

Learn more about work:

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