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A block of mass 10 kg slides down an inclined plane at an angle of 30ยฐ. The inclined plane has no friction and the block starts at a height of 2 m. How much kinetic energy does the block have when it reaches the bottom? (Acceleration due to gravity is [tex]\( g = 9.8 \, \text{m/s}^2 \)[/tex]).

A. 100 J
B. 0 J
C. 196 J
D. 20 J



Answer :

GinnyAnswer
To find the kinetic energy of the block when it reaches the bottom of the inclined plane, we can use the concept of energy conservation. In the absence of friction, all the potential energy (PE) of the block at the top of the incline will convert into kinetic energy (KE) at the bottom.

Here are the step-by-step details:

1. Determine the mass (m) of the block:
[tex]\[ m = 10 \ \text{kg} \][/tex]

2. Determine the height (h) from which the block starts:
[tex]\[ h = 2 \ \text{meters} \][/tex]

3. Acceleration due to gravity (g) is given as:
[tex]\[ g = 9.8 \ \text{m/s}^2 \][/tex]

4. Calculate the potential energy (PE) at the top of the incline:
The potential energy is given by the formula:
[tex]\[ PE = mgh \][/tex]
Substituting the known values:
[tex]\[ PE = 10 \ \text{kg} \times 9.8 \ \text{m/s}^2 \times 2 \ \text{m} \][/tex]
[tex]\[ PE = 196 \ \text{J} \][/tex]

5. Convert the potential energy to kinetic energy:
Since there is no friction, all the potential energy will convert into kinetic energy when the block reaches the bottom of the incline. Therefore:
[tex]\[ KE = PE \][/tex]
[tex]\[ KE = 196 \ \text{J} \][/tex]

Thus, the kinetic energy of the block when it reaches the bottom of the inclined plane is:

C. [tex]\(196 \ \text{J}\)[/tex]

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