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Aleksandr-060686 [28]
3 years ago
6

A 430 g soccer ball lying on the ground is kicked and flies off at 25 m/s. If the duration of the impact was .01 s, what was the

average force acting on the ball?
Physics
1 answer:
maw [93]3 years ago
4 0
<h2>The average force acting on the ball is 1075 N in the direction of travel of ball.</h2>

Explanation:

Force is given by rate of change of momentum.

Mass of soccer ball = 430 g = 0.43 kg

Initial velocity = 0 m/s

Final velocity = 25 m/s

Change in momentum = 0.43 x 25 - 0.43 x 0

Change in momentum = 10.75 kg m/s

Time taken = 0.01 s

Rate of change of momentum = Change in momentum ÷ Time

Rate of change of momentum = 10.75 ÷ 0.01

Rate of change of momentum = 1075 N

Force = 1075 N

The average force acting on the ball is 1075 N in the direction of travel of ball.

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3 years ago
A horizontal force of 750 N is needed to overcome the force of static friction between a level floor and a 250-kg crate. What is
Aleksandr [31]

Answer:

The acceleration of the crate is 1.8 m/s² so the answer is a.

Explanation:

The very first thing you must do when solving this problem is to draw a free body diagram. (The body diagram is attached to this answer)

So once we got the free body diagram, we can analyze it and build our sum of forces in the x and y directions. Notice that according to the diagram, there are 4 forces to this problem, Normal (N), Weight (W), kinetic friction (fk) and the 750N force.

As one may see in the free body diagram, two of the forces are vertical forces: N and W, so we can use them to build a sum of forces:

Starting with the sum of forces in the y-direction, we get:

ΣF_{y}=0

We set the sum equal to zero because there is no movement in the y-direction, so the system is in vertical equilibrium.

so the sum will be:

N-W=0

when solving for N we get that:

N=W

where W is found by multiplying the mass of the crate by the acceleration of gravity:

N=250kg*9.8m/s²

N=2450N

Once we found the normal force, we can use it to find the kinetic friction which is given by the following formula:

f_{k}=Nμ

where μ is the kinetic friction coefficient.

So we get that the kinetic friction is:

f_{k}=2450N*0.12

so

f_{k}=294

With this information we can go ahead and find the sum of horizontal forces:

ΣF_{x}=ma

In this case the sum is equal to mass times acceleration because the crate is moving horizontally due to the action of a force, so it will have an acceleration.

so the sum of forces look like this:

750N-f_{k}=ma

so

750N-294N=(250kg)a

when solving for a we get:

a=\frac{759N-294N}{250kg}\\ \\a=1.8m/s^{2}

so the crate's acceleration is 1.82m/s².

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