Lots of points! Will mark brainliest!!
1. How much force will be needed to move a 110kg box across the floor if the coefficient of friction is
0.40?
2. If the coefficient of friction is lowered to 0.25, but the same amount of force was applied, what would the acceleration of the object in problem #1 be?
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Ответ:
1. 431.2newtons
2. 1.47meters per second squared
Explanation:
f=ma
force of friction=normal force×coefficient of friction
normal force = mass× gravity
1.
nf=110kg × 9.8m/s^2 = 1078newtons
f=nf×cf
f=1078×.4=431.2
1
f=1078×.25=269.5
431.2-269.5=161.7
161.7=110kg× a
a=1.47
Ответ:
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