Question

question 13 (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° above the horizontal. if the mass of the box and the books is 40.0
kg, what is the acceleration of the box?
1.5 m/s2
1.9 m/s2
2.0 m/s2
3.3 m/s2

Explanation:

Step1: Find horizontal component of force

The horizontal component of the applied force \( F \) is given by \( F_{x}=F\cos\theta \), where \( F = 100\space N \) and \( \theta=37^{\circ} \). We know that \( \cos37^{\circ}\approx0.8 \), so \( F_{x}=100\times0.8 = 80\space N \).

Step2: Apply Newton's second law

Newton's second law is \( F = ma \), where \( F \) is the net force, \( m \) is the mass, and \( a \) is the acceleration. The net force in the horizontal direction is \( F_{x} \) (since the floor is smooth, there is no friction). The mass \( m = 40.0\space kg \). So, \( a=\frac{F_{x}}{m}=\frac{80}{40.0}=2.0\space m/s^{2} \)? Wait, no, wait. Wait, \( \cos37^{\circ} \) is approximately 0.8? Wait, no, actually, \( \cos37^{\circ}\approx0.8 \) (since \( \sin37^{\circ}\approx0.6 \), \( \cos37^{\circ}\approx0.8 \)). Wait, but let's recalculate. Wait, \( F = 100\space N \), angle \( 37^{\circ} \), so horizontal component \( F_{x}=F\cos\theta = 100\times\cos37^{\circ} \). If \( \cos37^{\circ}\approx0.8 \), then \( F_{x}=80\space N \). Then mass is 40 kg, so acceleration \( a=\frac{80}{40}=2.0\space m/s^{2} \)? But wait, maybe I made a mistake. Wait, no, wait, the options have 2.0 m/s² as an option. Wait, but let's check again. Wait, the force is applied at an angle above horizontal, so the horizontal component is \( F\cos\theta \), vertical component is \( F\sin\theta \), but since the floor is smooth, friction is zero, so the net force is the horizontal component. So \( F_{net}=F\cos\theta \). Then \( a = F_{net}/m=(F\cos\theta)/m \). Plugging in the numbers: \( F = 100\space N \), \( \theta = 37^{\circ} \), \( \cos37^{\circ}\approx0.8 \), \( m = 40\space kg \). So \( a=(100\times0.8)/40 = 80/40 = 2.0\space m/s^{2} \). Wait, but the options have 2.0 m/s². But wait, maybe my approximation of \( \cos37^{\circ} \) is wrong? Wait, actually, \( \cos37^{\circ}\approx0.7986\approx0.8 \), so that's correct. So the acceleration is \( 2.0\space m/s^{2} \)? But wait, the options include 2.0 m/s². Wait, but let me check again. Wait, maybe I messed up the angle. Wait, no, the problem says 37 degrees above horizontal. So horizontal component is \( F\cos\theta \). So yes, \( 100\times\cos37^{\circ}\approx80\space N \), divided by 40 kg is 2.0 m/s². But wait, the options have 2.0 m/s² as an option. Wait, but let me check with more precise value. \( \cos37^{\circ}\approx0.7986 \), so \( F_{x}=100\times0.7986\approx79.86\space N \). Then \( a = 79.86/40\approx1.9965\approx2.0\space m/s^{2} \). So the answer should be 2.0 m/s². Wait, but the options have 2.0 m/s². So that's the correct answer.

Answer:

2.0 m/s² (corresponding to the option with text "2.0 m/s2")