Chapter 7: Q. 24 (page 178)

The $1.0 k g$ block in FIGURE EX $7.24$ is tied to the wall with a rope. It sits on top of the $2.0 k g$ block. The lower block is pulled to the right with a tension force of $20 N$ . The coefficient of kinetic friction at both the lower and upper surfaces of the $2.0 k g$ block is $μ_{k} = 0.40$
a. What is the tension in the rope attached to the wall?
b. What is the acceleration of the $2.0 k g$ block?

Short Answer

Expert verified

(a) The tension in the rope attached to the wall is $3.92 N$

(b) The acceleration of the $2 k g$ block.

Step by step solution

Given information (part a)

Mass of lower block, $M = 2 k g$

Mass of the upper block, $m = 1 k g$

External force $F_{e x t} = 20 N$

Coefficient of kinetic friction $μ_{k} = 0.4$

Explanation (part a)

The tension in the rope will be equal to frictional force acting on 1 kg mass.

$T = μ_{k} m g m = 1 k g g = 9.8 m / s^{2} T = 0.4 (1 k g) (9.8 m / s^{2}) = 3.92 N$

The tension in the rope is $3.92 N$

Given information (part b)

Mass of lower block, $M = 2 k g$

Mass of the upper block, $m = 1 k g$

External force $F_{e x t} = 20 N$

Coefficient of kinetic friction $μ_{k} = 0.4$

Explanation (part b)

Frictional force at the upper and lower surface of 2 kg block will try to prevent its motion.

The equation of motion of 2 kg block is as follows.

$M a = F_{a} - T - f_{k} where f_{k} = μ_{k} (m + M) g M a = F_{a} - T - μ_{k} (m + M) g (2 k g) a = (20 N) - (3.92 N) - [(0.4) (1 k g + 2 k g) (9.8 m / s^{2})] (2 k g) a = 4.32 N a = \frac{4.32 N}{2 k g} a = 2.16 m / s^{2}$

The acceleration of the $2 k g$ block is $2.16 m / s^{2}$

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Short Answer

Step by step solution

Given information (part a)

Explanation (part a)

Given information (part b)

Explanation (part b)

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