(a) The work done by the force applied by the tractor is 79,968.47 J.
(b) The work done by the frictional force on the tractor is 55,977.93 J.
(c) The total work done by all the forces is 23,990.54 J.
<h3>
Work done by the applied force</h3>
The work done by the force applied by the tractor is calculated as follows;
W = Fd cosθ
W = (5000 x 20) x cos(36.9)
W = 79,968.47 J
<h3>Work done by frictional force</h3>
W = Ffd cosθ
W = (3500 x 20) x cos(36.9)
W = 55,977.93 J
<h3>Net work done by all the forces on the tractor</h3>
W(net) = work done by applied force - work done by friction force
W(net) = 79,968.47 J - 55,977.93 J
W(net) = 23,990.54 J
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Answer:
(A) 667.5 N/m
(B)
Explanation:
(A) Let the spring constant be k.
Using the formula F = kx
k = 251 / 0.376
K = 667.5 N/m
(B)
Work done
W = 0.5 × kx^2
W = 0.5 × 667.5 × 0.376 × 0.376
W = 47.2 J
Answer:
d. decreases
Explanation:
The law of conservation of momentum tells us that the sum of momenta before the collision is equal to the sum of momenta after the collision. The bag has no momentum as it falls onto the boat because its velocity is zero in the horizontal direction. But after it hits the boat, it's momentum increases while the momentum of the system remains the same. That means a component of the system must decrease somewhere else. And that component is the velocity, not the mass, of the boat.
Answer:
80m/s
Explanation:
to find it you have to work it out by using the formula distance divided by speed to find time.
