The mass m=20 kg is being pulled at a force F =200 N being oriented at 0=60degre

Question

The mass m=20 kg is being pulled at a force F =200 N being oriented at 0=60degree with ground overcoming a frictional force P=10 N. How much work is done to 10m? A cut moving at a velocity runs out of gas at a height of It 5 m from ground and starts to coast down the hill and coasts up the other side of lull What maximum height //, it can climb? direction and hits another mass M. moving m x direction at velocity V_2. Alter collision, both mass stuck together and starts moving at velocity V i , v direction Write down the momentum equation. A box of mass 2 kg slides along a horizontal, frictionless surface with a speed of VO m/s. The lx encounters

Explanation / Answer

27)

force , F = 200 N

theta = 60 degree

work done in moving 10 m = F * d * cos(60)

work done in moving 10 m = 200 * 10 * cos(60)

work done in moving 10 m = 1000 J

the work done in moving 10 m is 1000 J

28)

initial speed of ball , u = 50 m/s

acceleration , a = - 10 m/s^2

Using first equation of motion

v = u + a * t

v = 50 - 4 * 10

v = 10 m/s

kinetic energgy of ball = 0.5 * m * 10^2

kinetic energgy of ball = 5 *m

29)

let the maximum height is H

Using conservation of energy

0.5 * m * v^2 = m * g * (H - h)

0.5 * 30^2 = 9.8 (H - 5)

solving

H = 50 m

it will climb 50 m on other side

30)

Usign conservation for moementum

V * (M1 + M2) = (M1 * V1 - M2 * V2)

M1 * V + M2 * V = (M1 * V1 - M2 * V2)

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