The box is kept in motion at constant velocity by a force of F=99 N. Constant velocity means there is no acceleration, so the resultant of the forces acting on the box is zero. Apart from the force F pushing the box, there is only another force acting on it in the horizontal direction: the frictional force
which acts in the opposite direction of the motion, so in the opposite direction of F.
Therefore, since the resultant of the two forces must be zero,
so
The frictional force can be rewritten as
where
,
. Re-arranging, we can solve this equation to find
, the coefficient of dynamic friction:
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The mass on the left has a downslope weight of
W1 = 3.5kg * 9.8m/s² * sin35º = 19.7 N
The mass on the right has a downslope weight of
W2 = 8kg * 9.8m/s² * sin35º = 45.0 N
The net is 25.3 N pulling downslope to the right.
(a) Therefore we need 25.3 N of friction force.
Ff = 25.3 N = µ(m1 + m2)gcosΘ = µ * 11.5kg * 9.8m/s² * cos35º
25.3N = µ * 92.3 N
µ = 0.274
(b) total mass is 11.5 kg, and the net force is 25.3 N, so
acceleration a = F / m = 25.3N / 11.5kg = 2.2 m/s²
tension T = 8kg * (9.8sin35 - 2.2)m/s² = 27 N
Check: T = 3.5kg * (9.8sin35 + 2.2)m/s² = 27 N √
hope this helps. :)
Work done in compressing spring = 40.833
It simply needs a three-inch compression by applying a compressive force to a 15-inch long spring. we know that F(x)=kx. To act in displacement X and the poles' values cannot be substituted. Consequently, there will be 4. The spring constant K is therefore equal to 5, so. Three fit within this slot. We consequently derive the compressing spring constant K from this. , which is 5, 3. Now change this formula to include the value of the compressing spring constant. Four f will therefore equal 5.3 x and vice versa. Integration of four factors equals W.
In order to ensure the limit and further all of this, we currently have the range 0–7. Thus, this compressing spring task is completed. W is equal to 40.833
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