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Whitepunk [10]
4 years ago
15

Two carts have a compressed spring between them and are initially at rest. One of the carts has total mass, including its conten

ts, of 5.0 kg, and the other has total mass of 3.0 kg. If the 3.0 kg cart is moving at 3.0 m/s after the spring is allowed to push the carts apart, what is the velocity of the 5.0 kg cart after release? The system is closed.A) -1.8 m/s B) -5.0 m/sC) -0.56 m/sD) -0.0 m/s
Physics
1 answer:
ladessa [460]4 years ago
3 0

Answer:

A) - 1.8 m/s

Explanation:

As we know that whole system is initially at rest and there is no external force on this system

So total momentum of the system must be conserved

so we will have

m_1v_1 + m_2v_2 = 0

now plug in all data into above equation

5(v) + 3(3)

5v = -9

v = -1.8 m/s

so correct answer is

A) - 1.8 m/s

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3 years ago
You push your couch a distance of 3.9 m across the living room floor with a horizontal force of 220.0 n. the force of friction i
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The general formula to calculate the work is:

W=Fd \cos \theta

where F is the force, d is the displacement of the couch, and \theta is the angle between the direction of the force and the displacement. Let's apply this formula to the different parts of the problem.


(a) Work done by you: in this case, the force applied is parallel to the displacement of the couch, so \theta=0^{\circ} and \cos \theta=1, therefore the work is just equal to the product between the horizontal force you apply to push the couch and the distance the couch has been moved:

W=Fd=(220.0 N)(3.9 m)=858 J


(b) work done by the frictional force: the frictional force has opposite direction to the displacement, therefore \theta=180^{\circ} and \cos \theta=-1. Therefore, we must include a negative sign when we calculate the work done by the frictional force:

W=-Fd=-(144.0 N)(3.9 m)=-561.6 J


(c) The work done by gravity is zero. In fact, gravity (which points downwards) is perpendicular to the displacement of the couch (which is horizontal), therefore \theta=90^{\circ} and \cos \theta=0: this means

W=0.


(d) Work done by the net force:

The net force is the difference between the horizontal force applied by you and the frictional force:

F=220 N-144 N=76 N

And the net force is in the same direction of the displacement, so \theta=0^{\circ} and \cos \theta=1 and the work done is

W=Fd=(76 N)(3.9 m)=296.4 J


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