A 140 kg load is attached to a crane, which moves the load vertically. Calculate the tension in the cable for the following case: How far will the load move if it starts from rest and accelerates for 20 seconds?

The answer is below

Explanation:

A 140 kg load is attached to a crane, which moves the load vertically. Calculate the tension in the

cable for the following cases:

a. The load moves downward at a constant velocity

b. The load accelerates downward at a rate 0.4 m/s??

C. The load accelerates upward at a rate 0.4 m/s??

Solution:

Acceleration due to gravity (g) = 10 m/s²

a) Given that the mass of the crane (m) is 140 kg. If the load moves downward, the tension (T) is given by:

mg - T = ma

Since the load has a constant velocity, hence acceleration (a) = 0. Therefore:

mg - T = m(0)

mg - T = 0

T = mg

T = 140(10) = 1400 N

T = 1400 N

b)  If the load moves downward, the tension (T) is given by:

mg - T = ma

T = mg - ma = m(g - a)

T = 140(10 - 0.4) = 140(9.96) = 134.4

T = 134.4 N

c)  If the load moves upward, the tension (T) is given by:

T - mg = ma

T = ma + mg = m(a + g)

T = 140(0.4 + 10) = 140(10.4)

T = 145.6 N

2) To find the distance (s) if the load move from rest (u= 0) and accelerates for 20 seconds (t = 20). We use:

s = ut + (1/2)gt²

s = 0(20) + (1/2)(10)(20)²

s = 2000 m

speed

ratio of distance traveled to time interval

net force

vector sum of forces on object

object

source of diverging light rays either luminous or illuminated

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explanation: