A block of mass m309 g is dragged with a string across a rough horizontal table.

Question

A block of mass m309 g is dragged with a string across a rough horizontal table. The string tension is T= 3.43 N, and it pulls upward at an angle of 43.0° with the horizontal. At one particular instant the block is moving at a speed of v = 5.70 m/s. The coefficient of kinetic friction between the block and the table is 0.654. What is the power supplied to the block by the string tension? Number What is the power supplied by the force of friction? T-343 N Number v = 5.70 m/ s At this instant the block's speed is = 43.0 O decreasing O constant. 309 g increasing. O impossible to determine.

Explanation / Answer

We first need to find the force of friction.
Ffriction= coeff of friction x Normal reaction

Normal reaction + T sin(43) - mg = 0 (no motion in the y direction)
Normal reaction = mg- Tsin(43)
Normal reaction = 0.309*9.81 - 3.43*0.681= 0.6954 N
Ffriction= 0.654 * 0.6954 = 0.454 N

Power supplied to the block by the string tension = (Force due to tension- Ffriction)*velocity
= (3.43cos(43)-0.454)*5.70 = 11.71 N

Power by the force of friction = Ffriction* velocity = 0.454*5.70 = 2.5878 N

The velocity is increasing at this instance because power due to tension of string is greater than that of friction.

The energy E in kilowatt-hours (kWh) is equal to the power P in watts (W),

times the time period t in hours (hr) divided by 1000:

E(kWh) = P(W) × t(hr) / 1000

Here, Pw is 0.32 MW. Substituting the values we get the answer to be 2803200 kWH

Number of houses = 2803200/9500 ~= 295 houses. Option of 3 x 10^2 should be the closest one here.

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comment and I'll surely clarify your doubts. All the best!

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