An object of mass 10kg is drawn from height of 10cm. Find (g=10m/s) its kinetic energy on reading the ground.. Velocity before as it reaches the ground.
PE = mgh = 10*9.8*0.1 = 9.8J
all the PE is changed to KE, so
1/2 mv^2 = 9.8
Plug in m=10 and find v.
To find the kinetic energy of the object when it reaches the ground, we first need to find its potential energy at the initial height.
The potential energy (PE) of an object is given by the formula: PE = m * g * h, where m is the mass of the object, g is the acceleration due to gravity, and h is the height from which the object is dropped.
In this case, the mass of the object (m) is given as 10kg, the height (h) is 10cm (or 0.1m), and the acceleration due to gravity (g) is 10m/s^2.
So, the potential energy of the object at the initial height is:
PE = m * g * h = 10kg * 10m/s^2 * 0.1m
= 10J (Joules)
Now, when the object reaches the ground, all its potential energy is converted into kinetic energy (KE). The kinetic energy of an object is given by the formula: KE = 1/2 * m * v^2, where m is the mass of the object and v is its velocity.
So, we can equate the potential energy (PE) to the kinetic energy (KE) to find the velocity (v) at the ground:
PE = KE
10J = 1/2 * 10kg * v^2
Simplifying the equation:
10J = 5kg * v^2
2J = v^2
Taking the square root of both sides, we get:
√2J = v
Therefore, the velocity of the object just before it reaches the ground is approximately 1.414 m/s (rounded to three decimal places).