Question

the block shown in the figure is sliding along a frictionless horizontal surface. the blocks mass is m = 4.75 kg, the magnitude of the applied force is f = 30.0 n, and the angle of the applied force f below the horizontal is θ = 32.5°. determine the magnitude n of the normal force acting on the block. question 4.1c:

Explanation:

Step1: Analyze vertical - force components

In the vertical direction, the forces acting on the block are the gravitational force $F_g = mg$, the vertical component of the applied force $F_y=F\sin\theta$, and the normal force $N$. The net - force in the vertical direction is zero since there is no acceleration in the vertical direction ($a_y = 0$). According to Newton's second law $\sum F_y=0$. So, $N - mg - F\sin\theta=0$.

Step2: Solve for the normal force

We can re - arrange the equation $N - mg - F\sin\theta = 0$ to solve for $N$. We know that $m = 4.75$ kg, $g = 9.8$ m/s², $F = 30.0$ N, and $\theta=32.5^{\circ}$. First, calculate $mg$: $mg=4.75\times9.8 = 46.55$ N. Then, calculate $F\sin\theta$: $F\sin\theta=30\times\sin(32.5^{\circ})\approx30\times0.537 = 16.11$ N. Now, substitute these values into the equation for $N$: $N=mg + F\sin\theta$. So, $N=46.55+16.11 = 62.66$ N.

Answer:

$62.66$