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Readme [11.4K]
2 years ago
5

As an object slides across a rough horizontal

Physics
1 answer:
tiny-mole [99]2 years ago
6 0

When the object slides across the rough surface some of its potential energy will be lost to friction.

<h3>Conservation of mechanical energy</h3>

The law of conservation of mechanical energy states that the total mechanical energy of an isolated system is always constant.

M.A = P.E + K.E

When the object slides across the rough surface, some of the potential energy of the object will be converted into kinetic energy while the remaining potential energy will be converted into thermal energy due to frictional force of the rough surface.

P.E = K.E + thermal energy

Learn more about conservation of energy here: brainly.com/question/166559

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The question is in the picture
Sedbober [7]

Answer:

e) 120m/s

Explanation:

When the ball reaches its highest point, its velocity becomes zero, meaning

v_0-gt = 0.

where v_0 is the initial velocity.

Solving for t we get

t = \dfrac{v_0}{g}

which is the time it takes the ball to reach the highest point.

Now, after the ball has reached its highest point, it turns around and falls downwards. After time t_0 since it had reached the highest point, the ball has traveled downwards and the velocity v_f it has gained is

v_f = gt_0,

and we are told that this is twice the initial velocity v_0; therefore,

v_f = 2v_0  = gt_0

which gives

t_0 = \dfrac{2v_0}{g}.

Thus, the total time taken to reach velocity 2v_0 is

t_{tot} = t+t_0 = \dfrac{v_0}{g}+\dfrac{2v_0}{g}

t_{tot} = \dfrac{3v_0}{g}.

This t_{tot}, we are told, is 36 seconds; therefore,

36= \dfrac{3v_0}{g},

and solving for v_0 we get:

v_0 = \dfrac{36g}{3}

v_0 = \dfrac{36s(10m/s^2)}{3}

\boxed{v_0 = 120m/s}

which from the options given is choice e.

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