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MissTica
3 years ago
7

A block moves along a table at a constant velocity and then rolls off the edge of the table. The forces that should be included

in a free body diagram for the ball as it falls to the ground are:
normal force from table
friction force from table
gravity
force of fall
Physics
1 answer:
Alla [95]3 years ago
3 0

Answer:

Gravity

Explanation:

When the ball is falling to the ground, there is only one force- gravity. In some cases, there may be air resistance, but that seems to be neglected here.

Normal force:

Cannot be included. Normal force is only applicable when object is on a surface, and it acts perpendicular to the surface. Since the ball is falling, there is no surface, and therefore no normal force. This question gives you unnecessary information, designed to trick you. Please remember when normal force is applicable.

Friction force:

Also only applicable when object is moving, and is on the table. Friction only applies when there is an applied force. There is no applied force when the ball is falling, so therefore no friction force.

Force of fall:

First of all, what is this? There is nothing called force of fall.

Gravity:

This is the only one that applies. Just draw a vector arrow from the bottom of the ball and label it mg (acceleration due to gravity).

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2 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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