on the pleck fce of manitude t.4 N pushes a block of mass 4,09 ko across a floor

Question

on the pleck fce of manitude t.4 N pushes a block of mass 4,09 ko across a floor where the cofficlent of kinetic friction is 0,559. (o) How much work s doe y t What is the increase in thermal energy of the floor? (c) What is the increase in the kinetic energy of the block? (a) Number (b) Number (c) Number is 0.559. (a) How much work is done by that applied force osiend forces slides through a displacement of 4.20 m across the floor? (b) During that displacement, the thermal energy of the block increases by 42.7 Units Units Units

Explanation / Answer

(a) Given mass of the block = 4.09 kg
So, block's weight = (4.09)*(9.81) = 40.1 N
µk = 0.559
So, friction force = (0.559)*(40.1) = 22.4 N
displacement = 4.20 m

Therefore, work done by applied force = 46.4*4.20 = 194.9 J

(c) Work done against friction = (22.4)*(4.20) = 94.1 J
Work that increases KE of block = 194.9 - 94.1 = 100.8 J

(b) Now, block's thermal energy(TE) increased by 42.7 J means

Floor's (TE) increase = 94.1 - 42.7 = 51.4 J

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