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expeople1 [14]
3 years ago
8

A cardboard box sits on top of a concrete sidewalk where the coefficient of friction between the surfaces is 0.4. The mass of th

e box is 10 Kg and the box is pulled horizontally with a force of 60 N. What is the Gravitational Force on the box? What is the Normal Force? What is the Frictional Force? What is the Acceleration?

Physics
1 answer:
Sloan [31]3 years ago
6 0

Answer:

Fg = 98.1 [N]; N = 98.1 [N]; Ff = 39.24 [N]; a = 2.076[m/^2]

Explanation:

To solve this problem, we must make a free body diagram and interpret each of the forces acting on the box. In the attached diagram we can find the free body diagram.

The gravitational force is equal to:

Fg = (10 * 9.81) = 98.1 [N]

Now by summing forces on the Y axis equal to zero, we can find the normal force exerted by the surface.

N - Fg = 0

N = Fg

N = 98.1 [N]

The friction force is defined as the product of normal force by the coefficient of friction.

Ff = N * μ

Ff = 98.1 * 0.4

Ff = 39.24 [N]

By the sum forces on the x-axis equal to the product of mass by acceleration (newton's second law), we can find the value of acceleration.

60 - Ff = m * a

60 - 39.24 = 10 * a

a = 2.076[m/^2]

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Answer:

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Explanation:

hope this helps

4 0
2 years ago
A car with a mass of 2.0 * 10^3 kg is traveling at 15 m/s. what is the momentum of the car?
Allushta [10]
Hello,


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3 years ago
if a pressure of 70.kPa on a volume of 80.cm cubed is reduced to 10.kPa, by what factor does the pressure change
boyakko [2]

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6 0
3 years ago
if abus travelling at 20m/s is subject to steady decceleration of 5m/s².how long will it take yo come to rest.​
Likurg_2 [28]

Answer:

4 seconds

Explanation:

Given:

v₀ = 20 m/s

v = 0 m/s

a = -5 m/s²

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6 0
3 years ago
Holly puts a box into the trunk of her car. Later, she drives around an unbanked curve that has a radius of 48 m. The speed of t
LiRa [457]

Answer:

The minimum coefficient of friction is 0.544

Solution:

As per the question:

Radius of the curve, R = 48 m

Speed of the car, v = 16 m/s

To calculate the minimum coefficient of static friction:

The centrifugal force on the box is in the outward direction and is given by:

F_{c} = \frac{mv^{2}}{R}  

f_{s} = \mu_{s}mg

where

\mu_{s} = coefficient of static friction

The net force on the box is zero, since, the box is stationary and is given by:

F_{net} = f_{s} - F_{c}  

0 = f_{s} - F_{c}  

\mu_{s}mg = \frac{mv^{2}}{R}  

\mu_{s} = \frac{v^{2}}{gR}  

\mu_{s} = \frac{16^{2}}{9.8\times 48} = 0.544  

3 0
3 years ago
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