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lisov135 [29]
3 years ago
15

A horizontal force of 750 N is needed to overcome the force of static friction between a level floor and a 250-kg crate. If g-9.

8 m/s, what is the coefficient of static friction? a. 3.0 b. 0.15 c. 0.28 d. 0.31
Physics
1 answer:
seraphim [82]3 years ago
7 0

Answer:

0.31

Explanation:

horizontal force, F = 750 N

mass of crate, m = 250 kg

g = 9.8 m/s^2

The friction force becomes applied force = 750 N

According to the laws of friction,

Friction force = μ x Normal reaction of the surface

here, μ be the coefficient of friction

750 = μ x m g

750 = μ x 250 x 9.8

μ = 0.31

Thus, the coefficient of static friction is 0.31.

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A 56 kg sprinter, starting from rest, runs 49 m in 7.0 s at constant acceleration.what is the sprinter's power output at 2.0 s,
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The sprinter is in uniform accelerated motion, and its initial velocity is zero, so the relationship betwen space (S) and time (t) is
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a= \frac{2S}{t^2} = \frac{2 \cdot 49 m}{(7.0 s)^2} =2.0 m/s^2
So now we can solve the 3 parts of the problem.

a) power output at t=2.0 s
The velocity at t=2.0 s is
v(t)=at=(2.0 m/s^2)(2.0 s)=4.0 m/s

the kinetic energy of the sprinter is
K= \frac{1}{2} mv^2= \frac{1}{2}(56 kg)(4.0 m/s)^2=448 J

and so the power output is
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b) power output at t=4.0s 
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and so the power output is
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c) Power output at t=6.0 s
The velocity at t=2.0 s is
v(t)=at=(2.0 m/s^2)(6.0 s)=12.0 m/s

the kinetic energy of the sprinter is
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and so the power output is
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