i have noo idea how to do this, please hellppppp
A 15.6 kg block is dragged over a rough, hor-
izontal surface by a constant force of 72.2 N
acting at an angle of angle 34.8� above the
horizontal. The block is displaced 24.6 m and
the coefficient of kinetic friction is 0.234.
Find the work done by the 72.2 N force.
The acceleration of gravity is 9.8 m/s2 .
Answer in units of J.
72.2 cos34.8* 24.6 m = ___ J
That is the work done no matter what the friction force is.
Friction determines how much of that work goes into heat and how much goes into kinetic energy.
To find the work done by the 72.2 N force, we need to calculate the force parallel to the displacement of the block.
First, we find the vertical component of the force by multiplying the force magnitude by the sine of the angle:
Vertical component = 72.2 N * sin(34.8º)
Next, we find the horizontal component of the force by multiplying the force magnitude by the cosine of the angle:
Horizontal component = 72.2 N * cos(34.8º)
Since the force is acting at an angle, only its horizontal component contributes to the work done on the block.
Now, we find the frictional force acting on the block using the formula:
Frictional force = coefficient of kinetic friction * normal force
The normal force is equal to the weight of the block, which can be calculated using the formula:
Weight = mass * acceleration due to gravity
Finally, we can calculate the work done by the 72.2 N force using the formula:
Work = force * distance
Please plug in the given values into the equations and calculate each step to find the final answer in units of J (Joules).