If you increase the mass of an object and want to move an object a specific distance, then you need to do extra work than the earlier
<h3>What is work done?</h3>
The total amount of energy transferred when a force is applied to move an object through some distance
Work Done = Force * Displacement
For example, let us suppose a force of 10 N is used to displace an object by a displacement of 5 m then the work done on the object can be calculated by the above-mentioned formula
work done = 10 N ×5 m
=50 N m
Thus, when an object's mass is increased and it is desired to move it a certain distance, more work must be done than previously.
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Answer:
D. 2.8 × 10⁹ N
Explanation:
The force between two charges is directly proportional to the amount of charges at the two points and inversely proportional to the square of distance between the two points.
Fe= k Q₁Q₂/r²
Q₁= -0.0045 C
Q₂= -0.0025 C
r= 0.0060 m
k= 9.00 × 10 ⁹ Nm²/C²
Fe= (9.00 × 10 ⁹ Nm²/C²×-0.0045 C×-0.0025 C)/0.0060²
=2.8 × 10⁹ N
45km/h * 0.5h= 22.5km
The car can travel 22.5km in 0.5 hours
Answer:
31677.2 lb
Explanation:
mass of hammer (m) = 3.7 lb
initial velocity (u) = 5.8 ft/s
final velocity (v) = 0
time (t) = 0.00068 s
acceleration due to gravity (g) 32 ft/s^{2}
force = m x ( a + g )
where
- m is the mass = 3.7 lb
- g is the acceleration due to gravity = 32 ft/s^{2}
- a is the acceleration of the hammer
from v = u + at
a = (v-u)/ t
a = (0-5.8)/0.00068 = -8529.4 ( the negative sign showa the its decelerating)
we can substitute all required values into force= m x (a+g)
force = 3.7 x (8529.4 + 32) = 31677.2 lb
To solve this problem it is necessary to apply the concepts related to Newton's second Law and the force of friction. According to Newton, the Force is defined as
F = ma
Where,
m= Mass
a = Acceleration
At the same time the frictional force can be defined as,

Where,
Frictional coefficient
N = Normal force (mass*gravity)
Our values are given as,

By condition of Balance the friction force must be equal to the total net force, that is to say



Re-arrange to find acceleration,



Therefore the acceleration the horse can give is 