Answer:
Explanation:
Since the sled plus passenger moves with constant velocity , force applied will be equal to frictional force. Let the force applied be F
a ) Frictional force = μ R = F cosφ
R = mg - F sinφ
μ(mg - F sinφ) = F cosφ
μmg = F (μsinφ+cosφ)
F = μmg / (μsinφ+cosφ)
Work done
= F cosφ x d
= μmg x cosφ x d / (μsinφ+cosφ)
b )Work done
= 0.13 x 52.3 x 9.8 cos36.7 x 21.8 / ( 0.13 sin36.7 +cos36.7)
= 1164.61 / .87946
1324.23 J
c ) work done on the sled by friction
= - (work done by force)
= - μmg x cosφ x d / (μsinφ+cosφ)
d ) work done on the sled by friction
= - 1324.23 J
Joule (J) is the metric unit for energy
Answer:
105.1N
Explanation:
According to the newton second law
Since the acceleration is zero, then;
Since;
Given the following
mass of ladder m = 14kg
acceleration due to gravity g = 9.8m/s²
θ = 50°
Substitute
Hence the magnitude of the friction force (in N) exerted on the ladder in the point of contact with the horizontal surface is 105.1N
Answer: 490m
Explanation: 1/2 * 9.8m/s/s * 10s
Wavelength= velocity/frequency
5/2.5=2
