A student is able to move a block of mass 2 kg with minimum force of 1N. The force of friction between the object and the surface will be
The Correct Answer and Explanation is:
To determine the force of friction between the object and the surface, we first need to understand the concept of friction and how it relates to the applied force. Friction is a resistive force that opposes the relative motion between two surfaces in contact. In this case, the student needs to apply a minimum force of 1 N to start moving the block. This means that the force of static friction is equal to the minimum applied force needed to overcome it.
Static Friction:
Static friction is the frictional force that resists the initiation of motion between two objects. It is proportional to the normal force (the force pressing the two surfaces together) and is given by the formula:fstatic=μsNf_{\text{static}} = \mu_{\text{s}} Nfstatic=μsN
Where:
- fstaticf_{\text{static}}fstatic is the static frictional force,
- μs\mu_{\text{s}}μs is the coefficient of static friction,
- NNN is the normal force, which is the weight of the block in this case.
Since the block has a mass of 2 kg, its weight (normal force) is:N=mg=2 kg×9.8 m/s2=19.6 NN = mg = 2 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 19.6 \, \text{N}N=mg=2kg×9.8m/s2=19.6N
The minimum force required to overcome the static friction is given as 1 N. Therefore, the static frictional force at the point of motion must also be 1 N.
Thus, we can conclude that the force of static friction between the object and the surface is 1 N. The coefficient of static friction can be calculated as:μs=fstaticN=1 N19.6 N=0.051\mu_{\text{s}} = \frac{f_{\text{static}}}{N} = \frac{1 \, \text{N}}{19.6 \, \text{N}} = 0.051μs=Nfstatic=19.6N1N=0.051
So, the force of friction between the block and the surface is 1 N, and the coefficient of static friction is approximately 0.051. This is the minimum amount of force the student needs to apply to overcome friction and move the block.
