Question

equilibrium review
started: dec 10 at 7:53pm
quiz instructions
you can complete this review up to 3 times. it will be worth 2.5 points extra credit onto the exam.
question 6
1 pts
a student places an apple with a mass of 0.1 kg on a scale. the student gently lifts up on the apple. the scale reads 0.48 newtons. what must be the lifting force that the student has applied on the apple?

Explanation:

Step1: Calculate weight of apple

Weight \( W = mg \), where \( m = 0.1\,\text{kg} \), \( g = 9.8\,\text{m/s}^2 \).
\( W = 0.1 \times 9.8 = 0.98\,\text{N} \).

Step2: Analyze forces on apple

Forces on apple: weight (\( W \), downward), lifting force (\( F \), upward), scale force (\( N = 0.48\,\text{N} \), upward).
Equilibrium: \( F + N = W \).

Step3: Solve for lifting force \( F \)

Rearrange: \( F = W - N \).
Substitute values: \( F = 0.98 - 0.48 = 0.5\,\text{N} \).

Answer:

\( 0.5\) Newtons