A 50.0-kg block is being pulled up a 16.0° slope by a force of 250 n that is parallel to the slope. the coefficient of kinetic
friction between the block and the slope is 0.200. what is the acceleration of the block
1 answer:
When dealing with multiple forces acting on a body, it is advisable to draw a free-body diagram like that shown in the picture. There are four forces acting on the box: weight (W) pointing straight down, normal force perpendicular to the slope denoted as Fn, force used to push the box upwards along the slope and the frictional force acting opposite to the direction of motion of the box denoted as Ff. Frictional force is equal to coefficient of kinetic friction (μk) multiplied with Fn.
∑Fy = Fn - mgcos30° = 0
Fn = (50)(9.81)(cos 16) = 471.5 N
When in motion, the net force is equal to mass times acceleration according to Newton's 2nd Law of Motion:
Fnet = F - μk*Fn - mgsin30° = ma
250 - (0.2)(471.5 N) - (50)(sin 16°) = (50)(a)
a = 2.84 m/s²
∑Fy = Fn - mgcos30° = 0
Fn = (50)(9.81)(cos 16) = 471.5 N
When in motion, the net force is equal to mass times acceleration according to Newton's 2nd Law of Motion:
Fnet = F - μk*Fn - mgsin30° = ma
250 - (0.2)(471.5 N) - (50)(sin 16°) = (50)(a)
a = 2.84 m/s²
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Answer:
a)F=698.83 N
b)K=8221.56 N/m
Explanation:
Given that
mass ,m = 0.12 kg
Amplitude ,A= 8.5 cm
time period ,T = 0.2 s
We know that
We know that
K=Spring constant
K=8221.56 N/m
The maximum force F
F= K A
F= 8221.56 x 0.085 N
F=698.83 N
a)F=698.83 N
b)K=8221.56 N/m
Answer:
0.0059
Explanation:
According to the question the seismic activity density is given by
Here,
Number of Earthquakes over a given time span = 424
The land area affected = 71300 mi²
So,
The seismic activity density is 0.0059