Question

additional problems question 4.1a: the block shown in the figure is sliding along a frictionless horizontal surface the blocks mass is m = 5.45 kg, the magnitude of the applied force is f = 21.5 n, and the angle of the applied force f from the horizontal is θ = 28.0°. determine the magnitude n of the normal force acting on the block. n question 4.1b:

Explanation:

Step1: Analyze vertical - force equilibrium

In the vertical direction, the sum of forces is zero since there is no acceleration in the vertical direction ($a_y = 0$). The forces acting in the vertical direction 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 equation for vertical - force equilibrium is $\sum F_y=n - mg+F\sin\theta = 0$.

Step2: Solve for the normal force $n$

We can re - arrange the equation $\sum F_y = 0$ to solve for $n$. So, $n=mg - F\sin\theta$.
We know that $m = 5.45$ kg, $g = 9.8$ m/s², $F = 21.5$ N, and $\theta=28.0^{\circ}$.
First, calculate $F\sin\theta$: $F\sin\theta=21.5\times\sin(28.0^{\circ})=21.5\times0.4695\approx10.1$ N.
Then, calculate $mg$: $mg = 5.45\times9.8=53.41$ N.
Finally, $n=mg - F\sin\theta=53.41-10.1 = 43.31$ N.

Answer:

$43.3$ N