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attashe74 [19]
3 years ago
6

A person pushes horizontally with a force of 220. N on a 61.0 kg crate to move it across a level floor. The coefficient of kinet

ic friction is 0.270. (a) What is the magnitude of the frictional force? (b) What is the magnitude of the crate's acceleration? Use g=9.81 m/s2.
Physics
1 answer:
Ede4ka [16]3 years ago
7 0

Answer:

(a) 161.57 N

(b) 0.958 m/s^2

Explanation:

Force applied, F = 220 N

mass of crate, m = 61 kg

μ = 0.27

(a) The magnitude of the frictional force,

f = μ N

where, N is the normal reaction

N = m x g = 61 x 9.81 = 598.41 N

So, the frictional force, f = 0.27 x 598.41

f = 161.57 N

(b) Let a be the acceleration of the crate.

Fnet = F - f = 220 - 161.57

Fnet = 58.43 N

According to newton's second law

Fnet = mass x acceleration

58.43 = 61 x a

a = 0.958 m/s^2

Thus, the acceleration of the crate is 0.958 m/s^2.  

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Explanation:

The given data is as follows.

      F = 3.2 N,      m = 18.2 kg,

      t = 0.82 sec

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          I = Ft = \Delta P

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Putting the given values into the above formula as follows.

      v_{f} = \frac{Ft}{m} + v_{i}

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