Question

unit 2 exam mcq portion (10 question version)
a block of mass m is accelerated across a rough surface by a force of magnitude f that is exerted at an angle ⟨⟩ with the horizontal, as shown above. the frictional force on the block exerted by the surface has magnitude f.

1. what is the acceleration of the block?

(a) $\frac{f}{m}$
(b) $\frac{fcosphi}{m}$
(c) $\frac{f - f}{m}$
(d) $\frac{fcosphi - f}{m}$
(e) $\frac{fsinphi - mg}{m}$

1. what is the coefficient of friction between the block and the surface?

(a) $\frac{f}{mg}$
(b) $\frac{mg}{f}$
(c) $\frac{mg - fcosphi}{f}$
(d) $\frac{f}{mg - fcosphi}$
(e) $\frac{f}{mg - fsinphi}$
on a level horizontal table, a toy race car of mass m moves with constant speed v around a flat circular racetrack of radius r.

1. which of the following best represents the minimum coefficient of static friction required for the race car to continue to follow the circular path shown?

Explanation:

Step1: Analyze horizontal forces on block

The horizontal component of force $F$ is $F\cos\phi$, and the frictional force is $f$ acting in the opposite - direction. According to Newton's second law $F_{net}=ma$. The net horizontal force $F_{net}=F\cos\phi - f$.

Step2: Solve for acceleration

Since $F_{net}=ma$, then $a=\frac{F_{net}}{m}=\frac{F\cos\phi - f}{m}$.

Step3: Analyze normal force on block

In the vertical direction, the forces acting on the block are the vertical component of $F$ ($F\sin\phi$) acting upwards, the weight $mg$ acting downwards, and the normal force $N$ acting upwards. The net vertical force is zero since there is no acceleration in the vertical direction ($a_y = 0$). So, $N+F\sin\phi=mg$, and $N = mg - F\sin\phi$.

Step4: Calculate coefficient of friction

The frictional force $f=\mu N$, where $\mu$ is the coefficient of friction. Then $\mu=\frac{f}{N}=\frac{f}{mg - F\sin\phi}$.

Answer:

1. D. $\frac{F\cos\phi - f}{m}$
2. E. $\frac{f}{mg - F\sin\phi}$