Mike pulls a 4.5-kg sled across level snow with a force of 22.5 N along a rope that is 35 degrees above the horizontal. If the sled moves a distance of 65.3 m, how much work does Mike do?

W = 1203.54 J = 1.2 KJ

Explanation:

The work done on an object is defined as the product of the displacement of the object and the force applied in the direction of that displacement. Hence, the formula for work done is as follows:

where,

W = Work done by Mike = ?

F = Force applied by Mike = 22.5 N

θ = Angle between force and displacement = 35°

d = displacement of sled = 65.3 m

Therefore,

W = 1203.54 J = 1.2 KJ

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft

just took the test its not "d"