Question

1. a block weighing 15 newtons is pulled to the top of an incline that is 0.20 meter above the ground, as shown below. if 4.0 joules of work are needed to pull the block the full length of the incline, how much work is done against friction? explain.

Explanation:

Step1: Calculate work done against gravity

The work done against gravity is given by the formula \( W_{gravity} = F \times d \), where \( F \) is the weight of the block (15 N) and \( d \) is the vertical distance (0.20 m). So, \( W_{gravity} = 15 \, \text{N} \times 0.20 \, \text{m} = 3.0 \, \text{J} \).

Step2: Determine work done against friction

The total work done to pull the block (\( W_{total} = 4.0 \, \text{J} \)) is the sum of the work done against gravity (\( W_{gravity} \)) and the work done against friction (\( W_{friction} \)). So, \( W_{total} = W_{gravity} + W_{friction} \). Rearranging for \( W_{friction} \), we get \( W_{friction} = W_{total} - W_{gravity} \). Substituting the values, \( W_{friction} = 4.0 \, \text{J} - 3.0 \, \text{J} = 1.0 \, \text{J} \).

Answer:

The work done against friction is \( 1.0 \) joules.