The mass m = 20 kg is being pulled at a force F = 200 N being oriented \' Wl,h g

Question

The mass m = 20 kg is being pulled at a force F = 200 N being oriented ' Wl,h grouiKl overcoming a frictional force P = 10 N. Mow much work In done to move 10 m? (A) None (It) 900 J (C) 2000 J (D) 1900 J (E) 1000 N An object of mass m kg is thrown upward with a velocity of 50m/s. What is the KI ol the hull after sec? g = 10m/s A car moving at a velocity K 30m/s, runs out of gun at a height of h=5 m from ground and starts to coast down the hill and coasts up the other side of hill What maximum height H, it can climb? Mass M_1 is moving at a velocity V| in +x direction and hits another mass M_2 moving in x direction at velocity V_2. After collision, both mass stuck together and starts moving at velocity V direction. Write down the momentum equation.

Explanation / Answer

27) net force is Fnet = F*cos(theta)-P = 200*cos(60) - 10 = 90 N

work done by the net force W = F*S*cos(thea) = 90*10 = 900 J

28) what is the KE of the ball after ---- seconds ....

Question is incomplete

29) using law of conservation of energy

m*g*h + (0.5*m*V^2) = m*g*H

m cancels on both sides

(9.81*5) +(0.5*30*30) = 9.81*H

H= 50 .87 m

so the answer is E) 50 m

30)33 m/s = 33*18/5 =118.8 km/h

31) using law of conservation of momentum

M1V1 - M2V2 = (M1+M2)*V

So the answer is C)

32) F*cos(30) = 10

F = 10/cos(30) = 10/0.866 = 11.55 N

33) using law of conservation of energy

0.5*m*v^2 = 0.5*k*x^2

x = sqrt(m/k)*v = sqrt(2/500)*3= 0.19 m

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