A 10-KG mass is lifted upward, by a light cable . what is the tension in the cable if the acceleration is (A) zero, (B) 6m/s2 upward , and (C) 6m/s2 downward

(a) 98 N

(b) 158 N

(c) 38 N

Explanation:

Part (a)

When the acceleration is 0 m/s², the net force on the mass is 0 N. Therefore, the tension force is equal to the weight force due to Newton's Second Law:

∑F_y = T - w = ma_y ∑F_y = T - w = m(0 m/s²)∑F_y = T - w = 0 ∑F_y = T = w

Since the tension in the cable and the weight of the mass are equal to each other, we can solve for the weight force of the mass by using:

w = mg w = (10 kg)(9.8 m/s²)w = 98 N

Since T = w, we can say that T = 98 N.  

Part (b)

Let's set the upwards direction to be positive and the downwards direction to be negative. We can use Newton's Second Law to solve for the tension in the cable if the acceleration is 6 m/s² upward:

∑F_y = T - w = ma_y∑F_y = T - mg = m(6 m/s²)∑F_y = T - mg = 6m

Plug the known values into the equation and solve for T.

T - mg = 6mT - (10 kg)(9.8 m/s²) = 6(10 kg) T - 98 = 60 T = 158 N

The tension in the cable if the acceleration is +6 m/s² is 158 N.

Part (c)

The process is the same, but this time acceleration is -6 m/s².

∑F_y = T - w = ma_y∑F_y = T - mg = m(-6 m/s²)∑F_y = T - mg = -6m  

Plug known values into the equation and solve for T.

T - mg = -6mT - (10 kg)(9.8 m/s²) = -6(10 kg) T - 98 = -60 T = 38 N  

The tension in the cable if the acceleration is -6 m/s² is 38 N.

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