Question

science grade 7 classwork
name
date 1/15/26
make free body diagrams and show work.
part i:

1. an object of mass 3 kg is held with a rope of negligible mass. the block is then pulled up with an acceleration of 4 m/s². what is the weight of the object?

3 n
9.81 n
29.43 n
12 n
3 kg × 9.81 = 29.43 n

1. a person is holding a rope which has a block with weight 5n attached to it in the bottom. the person is moving the rope upward with constant speed. what is true about the tension of the rope when moving up?

a. t > 5n
b. t < 5n
c. t = 5n
d. t = 0

Explanation:

Question 1

Step 1: Recall the formula for weight

Weight \( W \) is calculated using the formula \( W = m \times g \), where \( m \) is the mass of the object and \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).

Step 2: Substitute the given values

Given \( m = 3 \, \text{kg} \) and \( g = 9.81 \, \text{m/s}^2 \), we substitute these into the formula:
\( W = 3 \, \text{kg} \times 9.81 \, \text{m/s}^2 \)

Step 3: Calculate the weight

\( W = 29.43 \, \text{N} \)

When an object moves with a constant speed, its acceleration is zero (since acceleration is the rate of change of velocity). According to Newton's second law, the net force acting on the object is \( F_{\text{net}} = m \times a \). Since \( a = 0 \), the net force \( F_{\text{net}} = 0 \). The forces acting on the block are the tension \( T \) upward and the weight \( W = 5 \, \text{N} \) downward. So, \( F_{\text{net}} = T - W \). Since \( F_{\text{net}} = 0 \), we have \( T - 5 \, \text{N} = 0 \), which means \( T = 5 \, \text{N} \).

Answer:

The weight of the object is \( 29.43 \, \text{N} \).

Question 2