contestada

If a machine exerts a force of 250 N on an object and no work is done, what must have occurred?
A) The machine has broken
B) The object has not moved
C) The object has gained momentum
D) The object has moved, but then stopped

A prisoner pushes on the prison wall trying to escape. He is unable to get the wall to move. Describe the work done by the prisoner.
A) The prisoner does no work because he applied no force.
B) The prisoner does a lot of work because the wall is heavy
C) The prisoner does work because extends a lot of effort
D) The prisoner does no work because the wall goes no distance

A 3.0-kg cart rolls down a hill. Assume that the height of the cart changes from a maximum of 4.0 meters at the top of the hill to 0.0 meters at the bottom. The speed of the cart changes from 0.0 m/s to 3.0 m/s. How much kinetic energy does the cart have when it is half way down the hill?
A) 6.0J
B) 12.0J
C) 58.8J
D) 117.6J

Respuesta :

Answer:

1) There is no work done by the machine because

B) The object has not moved

2) There is no work done by the prisoner because

D) The prisoner does no work because the wall goes no distance

3) The kinetic energy when it is half the way down is

6.0 J

Explanation:

1) As we know that the work done is the product of force and displacement

It is given as

[tex]W = Fdcos\theta[/tex]

so if the object is not displaced due to the force exerted by the object then the work done by the object must be ZERO

so correct answer is

B) The object has not moved

2) As we know that the work done is the product of force and displacement

It is given as

[tex]W = Fdcos\theta[/tex]

As we know that the wall is not displaced due to applied force so here work done by the prisoner must be zero

D) The prisoner does no work because the wall goes no distance

3) As we know by work energy theorem that work done by all forces is equal to change in its kinetic energy

So we will have

[tex]W_g + W_f = \frac{1}{2} mv^2[/tex]

so we will have

[tex]3(10)(4) + W_f = \frac{1}{2}(3)(3)^2[/tex]

[tex]120 + W_f = 13.5[/tex]

[tex]W_f = -106.5 J[/tex]

now when cart moves half the distance then again using the same

[tex]W_g + W_f = K[/tex]

[tex]K = 3(10)(2) - \frac{106.5}{2}[/tex]

[tex]K = 6.5 J[/tex]