Question

Can you answer in 30 min

Answer 1

The magnitude of the frictional force is (E).
`\mu_k` (mg+Tsinθ). Option E is the right choice.

The frictional force is always opposite to the direction of motion. In this case, the block is moving horizontally, so the frictional force is directed to the left.

The magnitude of the frictional force is given by:

`f = \mu_k N`

where
`\mu_k` is the coefficient of kinetic friction and N is the normal force.

The normal force is the force that the surface exerts on the block perpendicular to the surface. In this case, the normal force is equal to the weight of the block:

`N = mg`

The tension in the string, T, is acting at an angle θ above the horizontal. The horizontal component of the tension force is equal to the force that is pulling the block forward:

`T\cos\theta = ma`

The vertical component of the tension force is balanced by the weight of the block:

`T\sin\theta = mg`

Now we can substitute these values into the expression for the frictional force:

`f = \mu_k(mg + T\sin\theta)`

Therefore, the magnitude of the frictional force is (E).
`\mu_k` (mg+Tsinθ). Option E is the right choice.