A block of mass 2 kg is placed on the floor. The coefficient of static...
Weight of the block is 20N thus maximum friction that can act is 0.4 x 20 = 8N
But as the external force acting, F = 2.8N < 8N,
We get f = F = 2.8N
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A block of mass 2 kg is placed on the floor. The coefficient of static...
Given:
- Mass of block (m) = 2 kg
- Coefficient of static friction (μ) = 0.4
- Force applied (F) = 2.8 N
To find: Force of friction between block and floor
Calculations:
- The maximum force of static friction (Fs) can be found using the formula Fs = μN, where N is the normal force acting on the block.
- Since the block is placed on the floor, the normal force is equal to the weight of the block, which is given by W = mg, where g is the acceleration due to gravity (10 m/s2).
- Therefore, N = W = mg = 2 kg * 10 m/s2 = 20 N.
- The maximum force of static friction is Fs = μN = 0.4 * 20 N = 8 N.
- Since the applied force (F) is less than the maximum force of static friction (Fs), the block will not move and the force of friction will be equal to the applied force, i.e., f = F = 2.8 N.
Answer:
The force of friction between the block and floor is 2.8 N, which is option A.
A block of mass 2 kg is placed on the floor. The coefficient of static...
Friction is a self adjusting force. In this case force applied on block is less than limiting friction. i.e Friction, in this case,is Static friction..So to just stop the block from moving an equal opposing force is applied by floor surface..So Friction is 2.8 N (Option A).