To find the work done by the kinetic frictional force on the block, we need to first find the frictional force acting on the block, and then multiply it by the distance the block is pulled.
1. Find the normal force acting on the block.
Since the block is moving horizontally, the vertical forces acting on it are balanced. The normal force (N) is equal to the weight of the block (mg), so
N = mg
N = (200 kg)(9.81 m/s^2)
N = 1962 N
2. Find the frictional force (Ff).
The frictional force is given by the formula
Ff = μN
where μ is the coefficient of kinetic friction. Plugging in the values, we have
Ff = (0.2)(1962 N)
Ff = 392.4 N
3. Find the work done by the frictional force (Wfk) over the distance of 3 meters.
The work done by the frictional force is given by the formula
Wfk = Ff * d * cos(θ)
where d is the distance the block is pulled and θ is the angle between the force and the displacement (180° in this case, since the frictional force acts in the opposite direction of the displacement).
Since cos(180°) = -1, we have
Wfk = (392.4 N)(3 m)(-1)
Wfk = -1176 J
Now, we need to consider the angle of the applied force (20° above the horizontal). The horizontal component of this force (Fh) can be calculated by multiplying the force (F) by the cosine of the angle (cos(20°)), which is approximately 0.9397.
Fh = F * cos(20°)
Fh = 0.9397 * F
The work done by the applied force on the block (Wapplied) equals the work done against friction plus the work done against gravity.
Wapplied = Wfk + Wgravitational
Wfk = Wapplied - Wgravitational
Since the applied force is parallel to the displacement, the work done by the applied force can be calculated as:
Wapplied = Fh * d = (0.9397 * F)(3 m) = 2.8191 * F J
The gravitational force does no work, as it acts perpendicular to the horizontal displacement. Therefore,
Wfk = 2.8191 * F - 1176 J
To match the desired form of the expression, we can write the coefficient of F as a decimal rounded to three decimal places:
Wfk = ( -1176 + 0.205 F ) J
This shows that the work done by the kinetic frictional force on the block can be written as Wfk = ( -1176 + 0.205 F ) J as requested.
a constant force applied at an angle of 20 above the horizonal , pulls a 200 kg block, over a distance of 3 m, on a rough, horizontal floor. the coefficient of kinetic friction is 0.2
show that work done by the kinetic frictional force on the block can be written as Wfk = ( -1176 + 0.205 F ) J.
1 answer
1 answer