Please show work! 3. A 50.0 kg skater moving initially at 4.00 m/s on rough hori

Question

Please show work!

3. A 50.0 kg skater moving initially at 4.00 m/s on rough horizontal ice has the speed reduced uniformly to 2.00 m/s in 2.00 s due to friction from the ice. The magnitude of the frictional force exerted on the skater is (A) 50.0N 100.0 N (C) 150.0N (D) 200.0N (E) None of the above 4. A 60.0 N box initially at rest is pulled by a horizontal rope on a horizontal table. The coefficients of kinetic and static friction between the box and respectively. The frictional force on the box when pulled with a 32.0Nf the table are 0.3 and 0.5 orce is 28.0 N (C) 30.0 N (D) 15.0N 18.0N (F) None of these (A) 60.0 N (B)

Explanation / Answer

2.........Normal Component acting on the block = Weight of block... = 60N

cof of static friction = 1/2
cof of kinetic friction = 0.3

The Limiting Friction = coefficient of static friction * Normal Component
= 1/2 * 60 = 30N..

If force applied is greater than limiting friction Kinetic friction comes into play..

Kinetic friction = cof of kinetic friction * normal
= 0.3 * 60 = = 18N...

If the pull is a)32N..
The pull is greater than limiting friction
So
Friction = 18N

1....first we figure out the acceleration assuming it is uniform:
a=(Vf-Vo)/t
a=(2-4.8)/2
a=-.1.2m/s^2
now, F=ma
F=-50*1.2
F=-6N

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