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LiRa [457]
3 years ago
15

What is the value of work done on an object when a 50-newton force moves it 15 meters in the same direction as the force?

Physics
1 answer:
Volgvan3 years ago
4 0

Answer:

W = 750 Joules.

Explanation:

Given that,

Force acting on the object, F = 50 N

Distance moved by the object, d = 15 m

To find,

Work done by the object

Solution,

Let W is the work done by the object. It is given by the product of force and distance. Its expression is given by :

W=F\times d\ cos\theta

Here, \theta=0 (force and distance are in same direction)

W=F\times d

W=50\ N\times 15\ m

W = 750 Joules

So, the work done by the force is 750 Joules.

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You have a car (2000kg) that is accelerating uphill at 5.5 m/s what was the force (N) applied to the car? Remember (F=ma - mg
sp2606 [1]

Answer:

<em>The force applied to the car was 30,600 N</em>

Explanation:

According to the second Newton's law, the net force applied to an object of mass m is:

F_n=m.a\qquad\qquad [1]

Where a is the acceleration at which the object moves. The net force can be also calculated as the sum of all forces acting on the body.

We have a car of m=2000 Kg, being accelerated at 5.5 m/s^2 by a force F (unknown) directed upwards.

Considering the force is upwards and the weight of the car (W) is directed downwards, the net force is:

F_n=F-W\qquad\qquad [2]

Being W=m.g

Equating [1] and [2]:

F-W=m.a

Adding W:

F=W+m.a

F=m.g+m.a

F=m(g+a)

Substituting:

F=2000(9.8+5.5)

F=2000(15.3)

F=30,600 N

The force applied to the car was 30,600 N

5 0
3 years ago
Which force makes an object free fall?
LenaWriter [7]
Gravity

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How would a graph of velocity vs. Time would look like, if there was no motion at all? (Draw your graph)
Y_Kistochka [10]

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Explanation:

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The emission of electrons from a metal caused by light striking metal
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Two workers are sliding 300 kg crate across the floor. One worker pushes forward on the crate with a force of 400 N while the ot
almond37 [142]

Answer:

The kinetic coefficient of friction of the crate is 0.235.

Explanation:

As a first step, we need to construct a free body diagram for the crate, which is included below as attachment. Let supposed that forces exerted on the crate by both workers are in the positive direction. According to the Newton's First Law, a body is unable to change its state of motion when it is at rest or moves uniformly (at constant velocity). In consequence, magnitud of friction force must be equal to the sum of the two external forces. The equations of equilibrium of the crate are:

\Sigma F_{x} = P+T-\mu_{k}\cdot N = 0 (Ec. 1)

\Sigma F_{y} = N - W = 0 (Ec. 2)

Where:

P - Pushing force, measured in newtons.

T - Tension, measured in newtons.

\mu_{k} - Coefficient of kinetic friction, dimensionless.

N - Normal force, measured in newtons.

W - Weight of the crate, measured in newtons.

The system of equations is now reduced by algebraic means:

P+T -\mu_{k}\cdot W = 0

And we finally clear the coefficient of kinetic friction and apply the definition of weight:

\mu_{k} =\frac{P+T}{m\cdot g}

If we know that P = 400\,N, T = 290\,N, m = 300\,kg and g = 9.807\,\frac{m}{s^{2}}, then:

\mu_{k} = \frac{400\,N+290\,N}{(300\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}

\mu_{k} = 0.235

The kinetic coefficient of friction of the crate is 0.235.

5 0
3 years ago
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