An object, which is initially at rest on a frictionless horizontal surface, is a

Question

An object, which is initially at rest on a frictionless horizontal surface, is acted upon by four constant forces. F1 is 21.6 N acting due East, F2 is 41.2 N acting due North, F3 is 43.5 N acting due West, and F4 is 24.0 N acting due South. How much total work is done on the object in 4.44 seconds, if it has a mass of 12.0 kg?

An object, which is initially at rest on a frictionless horizontal surface, is acted upon by four constant forces F1 is 21.6 N acting due East, F2 is 41.2 N acting due North, F3 is 43.5 N acting due West, and F4 is 24.0 N acting due South. How much total work is done on the object in 4.44 seconds, if it has a mass of 12.0 kg? Number Units 612 Which type of energy is changing for the object while the above work is being done? O gravitational potential energy O internal energy O kinetic energy O elastic potential energy How fast does the object end up moving at the end of the 4.44 seconds? Number Units m/s 10.1 Incorrect.

Explanation / Answer

Given

Magnitude of Force 1 F1 = 21.6 N (east)

Magnitude of Force 2 F2 = 41.2 N (north)

Magnitude of Force 3 F3= 43.5 N (west)

Magnitude of Force 4 F4= 24.0 N (south)

Time t = 4.44 s

Mass of the object m = 12.0 Kg

Initial velocity u = 0 m/s (objet is at rest)

Solution

Four forces are acting on the object in all the four cardinal directions.

Forces acting in north and south are opposite to each other so the net force acting in north is the difference between them

Fnorth = F2 – F4

Fnorth = 41.2 – 24.0

Fnorth = 17.2 N

Forces acting in east and west are opposite to each other so the net force acting in north is the difference between them

Fwest = F3 – F1

Fwest = 43.5 – 21.6

Fwest = 21.9 N

Net force acting on the object is

F = ( Fnorth2 + Fwest2)

F = (17.22 + 21.92)

F = 27.8 N

Since this is a constant force the acceleration of the object is uniform

The acceleration

a = F/m

a = 27.8 /12.0

a = 2.32 m/s2

The distance moved in 4.44 s is

D = ut + ½ at2

D = (0 x 4.44) x + (½ x 2.32 x 4.442)

D = 22.9 m

Work done by the force

W = FD

W = 27.8 x 22.9

W = 636.62 J

W = 637 J (rounded off to three significant figures)

Work done on the object is 637 J

Since the work is done on the object its kinetic energy changes

Final speed of the object

v = u + at

v = 0 + 2.32 x 4.44

v = 10.3 m/s

Object will be moving at 10.3 m/s velocity at the end of 4.44 s

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