Question

In Figure, a block of mass m=5.00 kgis pulled along a horizontal frictionless floor by a cord that exerts a force of magnitude F=12.0Nat an angle$\mathit{\theta }\mathbf{=}\mathbf{25}\mathbf{.}\mathbf{0}\mathbf{°}$

(a) What is the magnitude of the block’s acceleration?

(b) The force magnitude Fis slowly increased. What is its value just before the block is lifted (completely) off the floor?

(c) What is the magnitude of the block’s acceleration just before it is lifted (completely) off the floor?

Short answer

a) The magnitude of the acceleration of the block is $2.18\mathrm{m}/{\mathrm{s}}^2$.

b) The magnitude of the force just before the block is lifted off the floor is 116N.

c) The magnitude of the acceleration just before the block is lifted off the floor is$21.0\mathrm{m}/{\mathrm{s}}^2$

Step-by-step solution

Given information

• The mass of the block $M=5\mathrm{kg}$.
• The force exerted on the block $F=12\mathrm{N}$.

Understanding the concept of Newton’s law

Newton’s second law states that the force acting on the body can be calculated with the product of mass and acceleration of the body.

The net force is equal to the vector sum of all the forces acting on the body, which helps us to understand the equilibrium conditions.

Using the Force body diagram and Newton’s second law, we can find the magnitude of the acceleration of the block and the magnitude of the force and acceleration just before the block is lifted off the floor.

Formulae:

Newton’s second law is,

$F_{net}=\sum _{}Ma$

(a) Calculate the magnitude of the block’s acceleration

The forces acting on the block are the weight of the block acting downward, the normal force from the floor on the block, and the force exerted by the cord. The free-body diagram can be used to understand the forces and their directions.

From FBD, we can write that,

$$\begin{gathered} F\cos \theta =Ma \\ a=\frac{\mathrm{Fcos\theta }}{\mathrm{M}} \\ =\frac{12\cos \left(25°\right)}{5} \\ =2.18\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

From FBD, we can write that,

$$\begin{gathered} F_N=Mg-F\sin \theta \\ =5\mathrm{kg}\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)-12\mathrm{Nsin}\left({25}^{°}\right) \\ =43\mathrm{N} \end{gathered}$$

As $F_N$is positive, the bock is on the floor when it moves with acceleration, $2.18\mathrm{m}/{\mathrm{s}}^2$.

(b) Calculate the value of F just before the block is lifted (completely) off the floor 

If the block is lifted off the floor, the normal force acting on it becomes zero. Therefore, we can write,

$$\begin{gathered} Mg=F\sin \theta \\ F=\frac{Mg}{\sin \theta } \\ =\frac{5\mathrm{kg}\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)}{\sin \left(25°\right)} \\ =116\mathrm{N} \end{gathered}$$

Therefore, the force on the box is 116N.

(c) Calculate the magnitude of the block’s acceleration just before it is lifted (completely) off the floor

As the box moves in the horizontal direction, the acceleration of the box is,

$$\begin{gathered} \mathrm{a}=\frac{\mathrm{Fcos\theta }}{\mathrm{M}} \\ =\frac{116\mathrm{N}.\cos \left(25°\right)}{5\mathrm{kg}} \\ =21\mathrm{m}/{\mathrm{s}}^{} \end{gathered}$$

Therefore, the acceleration of the box is $21\mathrm{m}/{\mathrm{s}}^2$.