Question

question 24
4 points
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(problem reference 4 - 2) a boy pulls a 5.00 - kg sled with a rope that makes a 60.0° angle with respect to the horizontal surface of a frozen pond. the boy pulls on the rope with a force of 10.0 n; and the sled moves with constant velocity.
what is the normal force acting on the sled?
o a. 49 n
o b. 40.3 n
o c. 57.7 n
o d. 98 n

Explanation:

Step1: Calculate the gravitational force

The gravitational force $F_g = mg$, where $m = 5.00\ kg$ and $g=9.8\ m/s^2$. So $F_g=5.00\times9.8 = 49\ N$.

Step2: Analyze the vertical - component of the pulling force

The pulling force is $F = 10.0\ N$ and the angle with the horizontal is $\theta = 60.0^{\circ}$. The vertical - component of the pulling force $F_{y}=F\sin\theta$. So $F_{y}=10.0\times\sin60^{\circ}=10.0\times\frac{\sqrt{3}}{2}\approx8.66\ N$.

Step3: Find the normal force

In the vertical direction, since the sled is in equilibrium ($a_y = 0$), $F_N+F_{y}-F_g = 0$. Then $F_N=F_g - F_{y}$. Substitute the values of $F_g$ and $F_{y}$: $F_N=49 - 8.66=40.34\ N\approx40.3\ N$.

Answer:

b. 40.3 N