# 1. A 10 kg block is hanging from a rope. Find the (magnitude of the) tension in the rope when i....

## Question:

1. A {eq}10 kg{/eq} block is hanging from a rope. Find the (magnitude of the) tension in the rope when

i. the block is stationary.

ii. the block is moving upward with a steady speed of {eq}5.0 m/s{/eq}

iii. the block is accelerating upward with a steady acceleration of {eq}5.0 m/s^2{/eq}

2. A {eq}1100 kg{/eq} car, whose wheels are locked, is being dragged across the pavement by a cable held at an angle of {eq}20^{\circ}{/eq} aboce the horizontal. The coefficient of kinetic friction between the car and the ground is {eq}\mu_k = 0.80{/eq}. The car starts from rest, and after {eq}3.0 s{/eq} is moving at a speed of {eq}1.5 m/s{/eq}

What is the (magnitude of the) tension in the cable {eq}T{/eq}?

## Newton's Second Law of Motion:

An object will accelerate when there is a net external force acting on that object. This is referred to as Newton's second law of motion. The acceleration is proportional to the force and inversely proportional to the mass of the object.

(1)

i. SInce the block is stationary, the tension is equal to the weight. Thus, we write:

{eq}T = mg \\ T = (10)(9.8) \\ T = 98 \ N {/eq}

(ii) The block is moving at a constant speed, thus the tension is still equal to the weight.

{eq}T = mg \\ T = (10)(9.8) \\ T = 98 \ N {/eq}

(iii) Since the block is accelerated upward, the apparent acceleration is given as:

{eq}g' = 9.8 - 5.0 \\ g' = 4.8 \ m/s^2 \\ T = (10)(4.8) \\ T = 48 \ N {/eq}

(2) First, we need to solve for the acceleration. We write:

{eq}a = \frac{v}{t} \\ a = \frac{1.5}{3} \\ a = 0.5 \ m/s^2 {/eq}

The sum of the horizontal force is given as:

{eq}T\cos{A} - umg = ma {/eq}

where:

• m is the mass of the car
• g is the acceleration due to gravity
• A is the angle
• a is the acceleration

Inserting the values of the available parameters, we write:

{eq}T\cos{20} - (0.8)(1100)(9.8) = (1100)(0.5) \\ T = 9762 \ N {/eq}