Answer: 3.21 N


For weight, we will multiply by 

Hence, the rock would weigh 3.21 N.
Answer:
Explanation:
During the first .8 s , the elevator is under acceleration . It starts from initial velocity u = 0 , final velocity v = 1.2 m /s , time = .8 s
v = u + at
1.2 = 0 + .8 a
a = 1.2 / .8
= 1.5 m /s²
During the acceleration in upward direction , let reaction force of ground on man be R .
Net force on man = R - mg
Applying Newton's 2 nd law
R - mg = ma
R = m ( g + a )
= 72 ( 9.8 + 1.5 )
= 813.6 N .
This reaction force will be measured by spring scale , so reading of spring scale will be 813.6 N .
Answer:
1977.696 J
Explanation:
Given;
Weight of the box = 28.0 kg
Force applied by the boy = 230 N
angle between the horizontal and the force = 35°
Therefore,
the horizontal component of the force = 230 × cosθ
= 230 × cos 35°
= 188.405 N
Coefficient of kinetic friction, μ = 0.24
Force by friction, f = μN
here,
N = Normal force = Mass × acceleration due to gravity
or
N = 28 × 9.81 = 274.68 N
therefore,
f = 0.24 × 274.68
or
f = 65.9232 N
Now,
work done by the boy, W₁ = 188.405 N × Displacement
= 188.405 N × 30
= 5652.15 J
and,
the
work done by the friction, W₂ = - 65.9232 N × Displacement
= - 65.9232 N × 30 m
= - 1977.696 J
[ since the friction force acts opposite to the direction of motion, therefore the workdone will be negative]
Answer:
1) 1.31 m/s2
2) 20.92 N
3) 8.53 m/s2
4) 1.76 m/s2
5) -8.53 m/s2
Explanation:
1) As the box does not slide, the acceleration of the box (relative to ground) is the same as acceleration of the truck, which goes from 0 to 17m/s in 13 s

2)According to Newton 2nd law, the static frictional force that acting on the box (so it goes along with the truck), is the product of its mass and acceleration

3) Let g = 9.81 m/s2. The maximum static friction that can hold the box is the product of its static coefficient and the normal force.

So the maximum acceleration on the block is

4)As the box slides, it is now subjected to kinetic friction, which is

So if the acceleration of the truck it at the point where the box starts to slide, the force that acting on it must be at 136.6 N too. So the horizontal net force would be 136.6 - 108.3 = 28.25N. And the acceleration is
28.25 / 16 = 1.76 m/s2
5) Same as number 3), the maximum deceleration the truck can have without the box sliding is -8.53 m/s2