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dybincka [34]
4 years ago
12

A block of mass 5 kg is placed on a rough table having a coefficient of static friction μs = 0.2.What is the force required to m

ove the block?
Physics
1 answer:
just olya [345]4 years ago
5 0

Answer:

<em>We need to (at least) apply a force of 9.8 N to move the block</em>

Explanation:

<u>Second Newton's Law</u>

If a net force F_n different from zero is applied to an object of mass m, then it will move at an acceleration a, given by

F_n=ma

If we apply a force F to an object placed on a rough surface, the only way to make it move is to beat the friction force  which is given by

F_r=\mu F_N

Where \mu is the static friction coefficient and F_N is the normal force exerted by the table to the object. Since there is no motion in the vertical direction the normal force equals the weight of the object:

F_N=mg=5\ kg\ 9.8\ m/s^2=49\ N

The friction force is

F_r=0.2 (49)=9.8\ N

Thus, we need to (at least) apply a force of 9.8 N to move the block

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6 0
4 years ago
A car speeds over a hill past point A, as shown in the figure. What is the maximum speed the car can have at point A such that i
Lubov Fominskaja [6]

Answer:

11.8 m/s

Explanation:

At the top of the hill, there are two forces on the car: weight force pulling down (towards the center of the circle), and normal force pushing up (away from the center of the circle).

Sum of forces in the centripetal direction:

∑F = ma

mg − N = m v²/r

At the maximum speed, the normal force is 0.

mg = m v²/r

g = v²/r

v = √(gr)

v = √(9.8 m/s² × 14.2 m)

v = 11.8 m/s

3 0
3 years ago
A train is travelling along a straight track at constant velocity from Western Station to Eastern station. The mile markers incr
S_A_V [24]

Answer:

Explanation:

Displacement of train = 60 - 25 = 35 mile

= 35 x 1.6 = 56 km

duration of time = 45 - 15 = 30 minutes

= 30 x 60 = 1800 s

velocity of train = displacement / time

= 56 / 1800 = .03111 km /s

= 31.111 m / s

3 0
3 years ago
Three 500-g point masses are at the corners of an equilateral triangle with 50-cm sides. What is the moment of inertia of this s
Ede4ka [16]

Answer:

0.25 kg m^2

Explanation:

mass of each , m = 500 g = 0.5 kg

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Moment of inertia about the axis passing through one corner and perpendicular to the plane of triangle

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I = 2 mr^2

I = 2 x 0.5 x 0.5 x 0.5

I = 0.25 kgm^2

6 0
3 years ago
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