A block of mass m = 2.90 kg is pushed a distance d = 5.80 m...

A block of mass m = 2.90 kg is pushed a distance d = 5.80 m along a frictionless horizontal table by a constant applied force of magnitude F = 16.0 N directed at an angle θ = 21.0° below the horizontal as shown in the figure below. (a) Determine the work done on the block by the applied force. J (b) Determine the work done on the block by the normal force exerted by the table. J (c) Determine the work done on the block by the force of gravity. J (d) Determine the work done by the net force on the block. J

Expert Solution

here the given data is force (F)= 16.0N

mass(M) = 2.90kg

distace moved(d) = 5.80m

and the angle () = 21.0⁰

a) work(W) = F*d

= Fcos*d

= 16(cos21)*5.80

= 86.304J

b) Since the object is not moving up or down, the force is negligible in those directions. Therefore zero work is done. Normal force is always perpendicular to the surface, which is in this case horizontal; this means the normal force is perfectly vertical. This means that there is no horizontal component of the force to affect the block's motion, so W=0N.

c) Same principle as above. Gravity works in the y direction, and it is not moving up or down, so there is no movement in the y direction.

d) Since gravity, the normal force, and the vertical component of the applied force cancel out vertically, the net force on the block is the horizontal component of the applied force. Thus, the net force on the block must, therefore, be the same as part a). The distance travelled does not change, so W miust also be the same.

genius_generous answered 1 week ago

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