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Sergio [31]
3 years ago
10

A force of 100.0N accelerates a 5.0 kg box at 15.0 m/s^2 along a level surface. Find the coefficient of sliding friction for the

box. After that, how far will the box have gone after 5.0 seconds of pulling it?
Physics
1 answer:
Phantasy [73]3 years ago
3 0

By balancing the force, we get :

100 - Frictional Force = ma

100 - μmg = ma

100 - μ(5)(10) = (5)(15)

50μ = 25

μ = 0.5  

Distance travelled by box in 5 seconds is :

d = ut + \dfrac{at^2}{2}\\\\d = 0 + \dfrac{15\times 5^2}{2}\\\\d = 187.5\ m

Hence, this is the required solution.

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5 0
3 years ago
A 27 kg bear slides, from rest, 14 m down a lodgepole pine tree, moving with a speed of 6.1 m/s just before hitting the ground.
Nadusha1986 [10]
<h2>Thus the force of friction is 235 N</h2>

Explanation:

When the bear was at the height of 14 m . Its potential energy = m g h

here m is the mass of bear , g is acceleration due to gravity and h is the height .

Thus P.E =  27 x 10 x 14 = 3780 J

The K.E of the bear just before hitting = \frac{1}{2} m v²

=   \frac{1}{2} x 27 x ( 6.1 )²  = 490 J

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Because the work done = Force x Distance

Thus frictional force = \frac{3290}{14} = 235 N

3 0
4 years ago
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