1. A 2.5 kg mass at rest on a horizontal surface is acted upon by an applied horizontal force of 150 N. A frictional force of 100 N acts on the mass. How much work is done on the mass?

my answer: would the work be 0?

2. When a rock is thrown straight up in the air, after it leaves the hand, the rock begins to slow down. This occurs because:

a) the rock is gaining potential energy as it rises and thus it must lose kinetic energy
b)the force of gravity acting on the rock increases as the rock rises
c)the forces acting on the rock are balanced
d)the potential energy of the rock decreases as the rock rises

3.4. Two objects have the same mass. One is travelling twice as fast as the other. The work that must be done to stop the faster object compared to the work required to stop the slower object is:

a)two times greater
b)the same
c)four times greater
d)half as great
e)one quarter as great

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2. 👎
3. 👁
1. 1. A 2.5 kg mass at rest on a horizontal surface is acted upon by an applied horizontal force of 150 N. A frictional force of 100 N acts on the mass. How much work is done on the mass?

my answer: would the work be 0?
-------------------
Yes, no motion, no work
-------------------

2. When a rock is thrown straight up in the air, after it leaves the hand, the rock begins to slow down. This occurs because:

a) the rock is gaining potential energy as it rises and thus it must lose kinetic energy
b)the force of gravity acting on the rock increases as the rock rises
c)the forces acting on the rock are balanced
d)the potential energy of the rock decreases as the rock rises

----------------------------
NO ! m g the same all the time
I say a) m g h + (1/2) m v^2 = constant
h goes up, v goes down

3.4. Two objects have the same mass. One is travelling twice as fast as the other. The work that must be done to stop the faster object compared to the work required to stop the slower object is:

a)two times greater
b)the same
c)four times greater
d)half as great
e)one quarter as great

NO !
(1/2) m v^2 and (1/2)m(2 v)^2
2^2 = 4
Ke of second four times the first
therefore 4 times as much work to stop
so (c)

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2. 👎

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