Ice fishing equipment weighing 3,000 N is pulled at a constant speed V = 0.5 m/s across a frozen lake by means of a
horizontal rope. If the coefficient of kinetic friction is 0.05, what is the work done (in J) by the fisherman in pulling the
equipment a distance of 500 m?
Just an explanation without numbers would be awesome, i just want to understand the correct steps to take.
Thanks!
W= (F(fr),s) = F(fr) •s•cosα.
α= 0, cosα= 1
F(fr)=μ•N=
W= μ•mg •s = 0.05•3000•500 =
thanks, but how come m is not in kg?
weight= mg = 3000 N
m= weight/g=3000/9.8 =306.12 kg
To find the work done by the fisherman in pulling the equipment across the frozen lake, we need to calculate the force exerted by the fisherman and the distance over which this force is applied.
First, let's calculate the force of kinetic friction acting against the equipment. The force of friction can be determined using the formula:
Frictional Force = μ * Normal Force
where μ is the coefficient of kinetic friction and the Normal Force is the force exerted by the surface perpendicular to the direction of motion. In this case, since the equipment is being pulled horizontally, the Normal Force is equal to the weight of the equipment.
Weight = Mass * Acceleration due to gravity
The weight of the equipment can be calculated by dividing its weight in Newtons by the acceleration due to gravity (9.8 m/s^2).
Weight = 3000 N / 9.8 m/s^2
Now we can calculate the force of kinetic friction:
Frictional Force = 0.05 * Weight
Next, we need to determine the total force applied by the fisherman. Since the equipment is being pulled at a constant speed, the force applied by the fisherman is equal in magnitude and opposite in direction to the force of kinetic friction.
Force Applied = - Frictional Force
Now we can calculate the work done by the fisherman using the formula:
Work = Force Applied * Distance
Work = (- Frictional Force) * Distance
Work = (-0.05 * Weight) * Distance
Finally, we can substitute the known values into the equation to find the work done by the fisherman:
Work = (-0.05 * (3000 N / 9.8 m/s^2)) * 500 m
Work = (-0.05 * 306.12 kg) * 500 m
Work = -765.3 J
Therefore, the work done by the fisherman in pulling the equipment a distance of 500 m is -765.3 Joules. The negative sign indicates that the work is done against the force of kinetic friction.