Asked by james
The magnitude of each force is 283 N, the
force on the right is applied at an angle 41
and the mass of the block is 27 kg. The
coefficient of friction is 0.283.
The acceleration of gravity is 9.8 m/s2 .
27 kg
μ = 0.283
283 N
41 degrees
283 N
What is the magnitude of the resulting acceleration?
Answer in units of m/s2
force on the right is applied at an angle 41
and the mass of the block is 27 kg. The
coefficient of friction is 0.283.
The acceleration of gravity is 9.8 m/s2 .
27 kg
μ = 0.283
283 N
41 degrees
283 N
What is the magnitude of the resulting acceleration?
Answer in units of m/s2
Answers
Answered by
Henry
Fr = 283N.[41o] + 283N.[0o]
X = 283*cos41 + 283 = 496.6 N.
Y = 283*sin41 = 185.7 N.
Fr^2 = X^2 + Y^2
Fr^2 = 496.6^2+185.7^2 = 281,096.05
Fr = 530.2 N.
Ff = u*mg = 0.283 * 264.6 = 74.9 N.
a = (Fr-Ff)/m=(530.2-74.9)/27=16.9 m/s^2
X = 283*cos41 + 283 = 496.6 N.
Y = 283*sin41 = 185.7 N.
Fr^2 = X^2 + Y^2
Fr^2 = 496.6^2+185.7^2 = 281,096.05
Fr = 530.2 N.
Ff = u*mg = 0.283 * 264.6 = 74.9 N.
a = (Fr-Ff)/m=(530.2-74.9)/27=16.9 m/s^2
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