91影视

• Mass of the block,$m=8.5\mathrm{kg}$.
• Angle of inclination of motion, $胃={30}^{掳}$.

From Newton鈥檚 second law, the net force acting on the body is equal to the vector sum of all the forces and is the product of the mass and acceleration of the body. The free-body diagram helps us to understand the forces acting on the body and their directions.

Using the free body diagram of the block, we can determine the tension in the block and the normal force acting on it.

Using Newton鈥檚 2^nd law, we can find the magnitude of the acceleration.

Formulae:

Force according to Newton鈥檚 second law,$F=ma$ (i)

Here, F is the force, mis the mass and ais the acceleration of the body.

Since the body is in equilibrium, the net force in the x direction on the body is zero. Write the all the forces in the x direction and substitute their values to calculate the tension as follows.

$$\begin{gathered} \underset{}{鈭�}{F}_{脳}=0 \\ T-mg\sin 胃=0 \\ (鈭�\mathrm{the}\mathrm{tension}\mathrm{in}\mathrm{the}\mathrm{cord}\mathrm{is}\mathrm{equal}\mathrm{to}\mathrm{the}\mathrm{sine}\mathrm{component} \\ \mathrm{of}\mathrm{the}\mathrm{weight}\mathrm{of}\mathrm{the}\mathrm{block},\mathrm{mg}) \\ T=mg\sin 胃 \\ =8.5\mathrm{kg}脳9.8\mathrm{m}/{\mathrm{s}}^2脳\sin {30}^{掳} \\ =41.65\mathrm{N} \\ 鈮�42\mathrm{N} \end{gathered}$$

So, the tension value is

Since the body is in equilibrium, the net force in the y direction on the body is zero. Write the all the forces in the y direction and substitute their values to calculate the normal force as follows.

$$\begin{gathered} \underset{}{鈭�}{F}_y=0 \\ {F}_N-mg\cos 胃=0 \\ (鈭�Thenormalforceisthe \\ perpendicularforceonthebook) \\ {F}_N=mg\cos 胃 \\ =8.5\mathrm{kg}脳9.8\mathrm{m}/{\mathrm{s}}^2脳\cos 30掳 \\ =72.05\mathrm{N} \\ 鈮�72\mathrm{N} \end{gathered}$$

Hence, the value of the normal force is$72\mathrm{N}$

As the cord is cut, there is no longer tension in the string i.e. T = 0 N Hence in this case,

The net force is given by equation (i). Use equation (i) and substitute the values of mass and net force to find the acceleration of the body.

$$\begin{gathered} F_{Net}=ma \\ -mg\sin 胃=ma(鈭�Fromfreebodydiagram) \\ a=-g\sin 胃 \\ =-9.8\mathrm{m}/{\mathrm{s}}^2脳\sin \left({30}^{掳}\right) \\ =-4.9\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Here the negative sign shows that the acceleration is acting down the plane and its value is$-4.9\mathrm{m}/{\mathrm{s}}^2$.