Question

question 8 (1 point)
a student pulls a box of books on a smooth horizontal floor with a force of 100 n in
a direction of 37.0° above the horizontal. if the mass of the box and the books is
40.0 kg, what is the normal force on the box?
292 n
312 n
332 n
392 n

Explanation:

Step1: Find the weight of the box

The weight \( W \) of an object is given by \( W = mg \), where \( m = 40.0\space kg \) and \( g = 9.8\space m/s^2 \).
\( W = 40.0\times9.8 = 392\space N \)

Step2: Find the vertical component of the pulling force

The pulling force \( F = 100\space N \) makes an angle \( \theta = 37.0^\circ \) with the horizontal. The vertical component \( F_y \) is \( F\sin\theta \).
\( F_y = 100\times\sin(37.0^\circ) \approx 100\times0.6 = 60\space N \) (using \( \sin(37^\circ)\approx0.6 \))

Step3: Apply equilibrium in the vertical direction

In the vertical direction, the normal force \( N \), the weight \( W \) (downward), and the vertical component of the pulling force \( F_y \) (upward) are in equilibrium. So, \( N + F_y = W \) (since \( F_y \) is upward, it reduces the normal force).
\( N = W - F_y \)
\( N = 392 - 60 = 332\space N \)

Answer:

332 N