A3.50 kg block is pushed along a horizontal floor by a force of magnitude 15.0 n at an angle θ = 30.0° with the horizontal. the coefficient of kinetic friction between the block and the floor is 0.250.
calculate the magnitudes of (a) the frictional force on the block from the floor and
(b) the block's acceleration.

Answer :

The frictional force on the block from the floor and the  block's acceleration are 10.45 N and 0.73 m/s².

Explanation :

Given that,

Mass of block = 3.50

Angle = 30°

Force = 15.0 N

Coefficient of kinetic friction = 0.250

We need to calculate the frictional force

Using formula of frictional force

(II). We need to calculate the block's acceleration

Using newton's second law of motion

Hence, The frictional force on the block from the floor and the  block's acceleration are 10.45 N and 0.73 m/s².

(a) The frictional force on the block is 10.45 N

(b) The acceleration of the block is 0.725 m/s²

The given parameters;

mass of the block, m = 3.5 kg

applied force, F = 15 N

angle of inclination θ = 30.0°

coefficient of kinetic friction, μ = 0.25

The resultant of vertical component of the for force on the block is calculated from Newton's second law of motion as shown below;

(a) The frictional force on the block is calculated as;

(b) The acceleration of the block is calculated as;

Learn more here:

spare some change and give it to the poor

explanation:

plz : ((