A)A block of mass 2.50 kg is pushed 2.40 malong a frictionless horizontal table
ID: 1764672 • Letter: A
Question
A)A block of mass 2.50 kg is pushed 2.40 malong a frictionless horizontal table by a constant 16.0 N force directed 25.0° below the horizontal. (1) Determine the work done by the appliedforce.1 J
(2) Determine the work done by the normal force exerted by thetable.
2 J
(3) Determine the work done by the force of gravity.
3 J
(4) Determine the work done by the net force on the block.
4 J
B)A 54 kg pole vaulter running at13 m/s vaults over the bar. Her speed whenshe is above the bar is 1.3 m/s. Neglectair resistance, as well as any energy absorbed by the pole, anddetermine her altitude as she crosses the bar.
1 m (1) Determine the work done by the appliedforce.
1 J
(2) Determine the work done by the normal force exerted by thetable.
2 J
(3) Determine the work done by the force of gravity.
3 J
(4) Determine the work done by the net force on the block.
4 J
B)A 54 kg pole vaulter running at13 m/s vaults over the bar. Her speed whenshe is above the bar is 1.3 m/s. Neglectair resistance, as well as any energy absorbed by the pole, anddetermine her altitude as she crosses the bar.
1 m
Explanation / Answer
mass M = 2.50 kgdistance S = 2.40 m force F = 16.0 N angle = 25.0° (1) the work done by the applied force =F cos * S = 34.8 J .
(2) the work done by the normal force exerted by the table =0 Since normal force and displacement areperpendicular to each other
(3) the work done by the force of gravity = 0 Since force of gravity and displacement areperpendicular to each other
(4) the work done by the net force on the block.= 34.8 + 0 +0 = 34.8 J
B)mass m = 54 kg initial speed v = 13 m/s Her speed when she is above the baris v' = 1.3 m/s kinetic energy K = ( 1/ 2) m v^ 2 = 4563J Total energy at maximum altitude K ' = (1/ 2) m v ' ^ 2 + mgh = 45.63 + 529.2 h K = K ' 45.63 + 529.2 h = 4563 h = 8.53 m
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