Question

1. a force of 10 n was used to lift an object to a height of 5.0 m. calculate the amount of work done. 4. a force of 3.0 n was used to move a barrel up an inclined plane. it was determined that 24 j of work was done. calculate the distance the barrel moved.

Explanation:

Problem 3:

Step1: Recall the work formula

The formula for work \( W \) done by a force \( F \) in lifting an object to a height \( h \) (when force is against gravity and displacement is vertical) is \( W = F \times h \) (since the force and displacement are in the same direction, \( \cos\theta = 1 \) where \( \theta \) is the angle between force and displacement).
Given \( F = 10\space N \) and \( h = 5.0\space m \).

Step2: Substitute values into the formula

Substitute \( F = 10\space N \) and \( h = 5.0\space m \) into \( W = F \times h \).
\( W = 10\space N \times 5.0\space m \)

Step3: Calculate the work

\( 10\times5.0 = 50 \), and the unit of work is Joule (J) since \( 1\space J = 1\space N\cdot m \). So \( W = 50\space J \).

Step1: Recall the work formula

The formula for work \( W \) is \( W = F \times d \), where \( F \) is the force applied and \( d \) is the distance moved in the direction of the force. We need to find \( d \), so we can rearrange the formula to \( d=\frac{W}{F} \).

Step2: Identify given values

Given \( F = 3.0\space N \) and \( W = 24\space J \).

Step3: Substitute values into the rearranged formula

Substitute \( W = 24\space J \) and \( F = 3.0\space N \) into \( d=\frac{W}{F} \).
\( d=\frac{24\space J}{3.0\space N} \)

Step4: Calculate the distance

\( \frac{24}{3.0}=8 \), and the unit of distance is meter (m) since \( 1\space J = 1\space N\cdot m \), so \( d = 8\space m \).

Answer:

The amount of work done is \( \boldsymbol{50\space J} \).

Problem 4: