91影视

According to Newton鈥檚 second law, the force acting on an object is directly related to the object鈥檚 mass and acceleration. For a higher-mass object, a higher value of force is required to accelerate it.

Part (a)

Here, $F$is the force that acts on the box, $N$is the normal force on the box, $m$is the mass of the box, $a$is the acceleration of the box, $胃$is the angle made by the force with the horizontal, $F\cos 胃$is the horizontal component of the force, and $F\sin 胃$is the vertical component of the force.

The expression for the force that acts on the box when it is pulled parallel to the horizontal can be written as

$F=ma鈥�\left(i\right)$

The horizontal force that acts on the box when the force is at inclination is given by

$F\cos 胃$.

The angle $\left(胃\right)$varies from 0 to 1 for cosine and sine angles. Then, the limits for $\cos 胃$will be

$0<\cos 胃<1$.

Multiplying the above relation with $F$,

$\begin{array}{c}F\left(0\right)<\left(F\right)\cos 胃<F\left(1\right)\\ 0<F\cos 胃<F\end{array}$.

Substituting the value of equation (i) in the above equation,

role="math" localid="1645082333072" $\begin{array}{c}0<F\cos 胃<ma\\ a>\frac{F\cos 胃}{m}鈥�\left(ii\right)\end{array}$

Consider that the acceleration of the box is $a\text{'}$when it is pulled at an inclination.

Then, the horizontal force that acts on the force will be

$F=ma\text{'}$.

Substituting this value in equation (ii),

$\begin{array}{c}a>\frac{ma\text{'}\cos 胃}{m}\\ a>a\text{'}\cos 胃\end{array}$.

Thus, from the above relation, it is clear that theacceleration of the box decreases if the box is pulled at an angle.

Part (b)

Here, $F$is the force that acts on the box, $N$is the normal force on the box, $m$is the mass of the box, $a$is the acceleration of the box, $胃$is the angle made by the tension force with the horizontal, $T\cos 胃$is the horizontal component of the tension force,$T\sin 胃$is the vertical component of the tension force, and $f$is the frictional force.

If there is no frictional force existing between the block and the table, the expression for the net force can be written as

$\begin{array}{c}F=ma\\ T=ma\\ a=\frac{T}{m}鈥�\left(iii\right)\end{array}$

Consider that $a\text{'}$is the acceleration of the box when there is a frictional force existing between the block and the table.

If there is a frictional force existing between the block and the table, the expression for the net force can be written as

role="math" localid="1645082487698" $\begin{array}{c}ma\text{'}=T\cos 胃-f\\ a\text{'}=\frac{T\cos 胃-f}{m}鈥�\left(iv\right)\end{array}$

The condition for $\cos 胃$is given by

$0<\cos 胃<1.$

The above expression for the tension force can be written as

$0<T\cos 胃<T$鈥� (v)

Thus, from equations (iii), (iv), and (v), it is clear that theacceleration of the box decreases if there is a frictional force existing between the table and the box.