Question #182132

A person is pushing a sled along a horizontal surface. They exert a force of 200 N on the 75 kg sled. In addition a frictional force of 20 N acts on the sled.

What is the rate of acceleration of the sled? (in m s−2 to 2.s.f)


Expert's answer

Solution.

F=200N;F=200N;

m=75kg;m=75kg;

Ff=20N;F_f=20N;

a−?;a-?;

ma→=F→+Ff→;\overrightarrow{ma}=\overrightarrow{F}+\overrightarrow{F_f};

ma=F−Ff  ⟹  a=F−Ffm;ma=F-F_f\implies a=\dfrac{F-F_f}{m};

a=200N−20N75kg=2.40ms−2;a=\dfrac{200N-20N}{75kg}=2.40ms^{-2};

Answer: a=2.40ms−2.a=2.40ms^{-2}.


LATEST TUTORIALS
APPROVED BY CLIENTS