First, determine the mass of the object by dividing its weight on Earth by 9.8 m/s² as shown below,
                                   m = 250 N / 9.8 m/s² = 25.51 kg
Then, multiply the obtained mass by the acceleration due to gravity (g) on Pluto. 
                                  W (in Pluto) = (25.51 kg) x (0.61 m/s²)  = 15.56 N
Therefore, the object will only weigh 15.56 N. 
        
             
        
        
        
Answer:
The ball fell 275.625 meters after 7.5 seconds
Explanation:
<u>Free fall
</u>
If an object is left on free air (no friction), it describes an accelerated motion in the vertical direction, powered exclusively by the acceleration of gravity. The formulas needed to compute the different magnitudes involved are


Where  is the final speed of the object in free fall, assumed positive downwards, t is the time elapsed since the release and y is the vertical distance traveled by the object
 is the final speed of the object in free fall, assumed positive downwards, t is the time elapsed since the release and y is the vertical distance traveled by the object
The ball was dropped from a cliff. We need to calculate the vertical distance the ball went down in t=7.5 seconds. We'll use the formula


 
 
        
             
        
        
        
Answer:
The sum of all forces for the two objects with force of friction F and tension T are:
(i) m₁a₁ = F
(ii) m₂a₂ = T - F
1) no sliding infers: a₁ = a₂= a
The two equations become:
m₂a = T - m₁a
Solving for a:
a = T / (m₁+m₂) = 2.1 m/s²
2) Using equation(i):
F = m₁a = 51.1 N
3) The maximum friction is given by:
F = μsm₁g 
Using equation(i) to find a₁ = a₂ = a:
a₁ = μs*g
Using equation(ii)
T = m₁μsg + m₂μsg = (m₁ + m₂)μsg = 851.6 N
4) The kinetic friction is given by: F = μkm₁g
Using equation (i) and the kinetic friction:
a₁ = μkg = 6.1 m/s²
5) Using equation(ii) and the kinetic friction:
m₂a₂ = T - μkm₁g 
a₂ = (T - μkm₁g)/m₂ = 12.1 m/s²