A block of mass m = 353 g is dragged with a string across a rough horizontal tab

Question

A block of mass m = 353 g is dragged with a string across a rough horizontal table. The string tension is T = 2.79 N, and it pulls upward at an angle of = 41.0° with the horizontal. At one particular instant the block is moving at a speed of v = 6.30 m/s. The coefficient of kinetic friction between the block and the table is k = 0.654.

A block of mass m = 353 g is dragged with a string across a rough horizontal table. The string tension is T-2.79 N, and it pullsupward at an angle of -41.0° with the horizontal. At one particular instant the block is moving at a speed of v = 6.30 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=279 N Number v=6.30 m/s At this instant the block's speed is =41.00 O decreasing O constant. O increasing 353 g impossible to determine.

Explanation / Answer

Given,

m = 353 g = 0.353 kg ; T = 2.79 N ; phi = 41 deg ; v = 6.3 m/s ; uk = 0.654

The vertical and horizontal component of tension force will be:

Ty = 2.79 sin41 = 1.83 N

Tx = 2.79 cos41 = 2.11 N

Power by tension

P(T) = Tx v = 2.11 x 6.3 = 13.3 W

Hence, P(T) = 13.3 W

W = mg = 0.353 x 9.81 = 3.46 N

Fnet = Tx - uk( W - Ty)

Fnet = 2.11 - 0.654 (3.46 - 1.83) = 1.044 N

So the power by friction is:

P(Ff) = P(T) - Fnet x v

P(Ff) = 1.044 x 6.3 - 13.3 = -6.72 W

Hence, P(Ff) = -6.72 W

a = Fnet/m = 0.947/.353 = 2.68 m/s^2

So its accelerating hence speed increasing.

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