Two blocks are connected by a very light string passing over a massless and frictionless pulley. The 20.0 N block moves 75.0cm to the right and the 12.0 N block moves 75.0cm downward.Find the total work done on 20.0N block if there is no friction between the table and the 20.0N block.

Answers

Answer 1

Answer: The string is assumed to be massless so the tension is the sting above the 12.0 N block has the same magnitude to the horizontal tension pulling to the right of the 20.0 N block. Thus,
1.22 a = 12.0 - T (eqn 1)
and for the 20.0 N block:
2.04 a = T - 20.0 x 0.325 (using µ(k) for the coefficient of friction)
2.04 a = T - 6.5 (eqn 2)

[eqn 1] + [eqn 2] → 3.26 a = 5.5
a = 1.69 m/s²

Subs a = 1.69 into [eqn 2] → 2.04 x 1.69 = T - 6.5
T = 9.95 N

Now want the resultant force acting on the 20.0 N block:
Resultant force acting on the 20.0 N block = 9.95 - 20.0 x 0.325 = 3.45 N
Units have to be consistent ... so have to convert 75.0 cm to m:

75.0 cm = 75.0 cm x [1 m / 100 cm] = 0.750 m
work done on the 20.0 N block = 3.45 x 0.750 = 2.59 J

Answer 2

Answer:

Final answer:

The total work done on the 20.0 N block as it moves 75.0 cm to the right, with no friction between the table and the block, is 15 Joules.

Explanation:

In physics, work done by a force is given by the formula Work = Force x Distance x Cos θ, where θ is the angle between the force and the direction of motion. Here, the force is the same as the weight of the 20.0 N block (because weight = mass * gravity, and the mass is the weight divided by gravity, so mass * gravity = weight), and because the block moves horizontally (rightward), θ is 0 degrees. The Cos of 0 degrees is 1.

So, the work done on the 20.0 N block as it moves 75.0 cm (or 0.75 meters, because 1 m = 100 cm) to the right is 20.0 N * 0.75 m * 1 = 15 Joules.

Learn more about Work Done here:

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