Question

for a moving object, the force acting on the object varies directly with the objects acceleration. when a force of 21 n acts on a certain object, the acceleration of the object is \\( \frac{3}{2} \frac{m}{s^2} \\). if the acceleration of the object becomes \\( 8 \frac{m}{s^2} \\), what is the force? \\( \square \\) n

Explanation:

Step1: Recall Direct Variation Formula

Force \( F \) varies directly with acceleration \( a \), so \( F = ka \), where \( k \) is the constant of variation. First, find \( k \) using the given \( F = 21 \, \text{N} \) and \( a=\frac{3}{2} \, \text{m/s}^2 \).

Substitute into \( F = ka \): \( 21 = k \times \frac{3}{2} \).

Solve for \( k \): \( k = 21 \div \frac{3}{2} = 21 \times \frac{2}{3} = 14 \).

Step2: Calculate New Force

Now, use \( k = 14 \) and the new acceleration \( a = \frac{8}{2} = 4 \, \text{m/s}^2 \) (simplify \( \frac{8}{2} \) to 4) in \( F = ka \).

Substitute \( k = 14 \) and \( a = 4 \): \( F = 14 \times 4 = 56 \).

Answer:

\( 56 \)