A 2.0-kg block is projected up a frictionless inclined plane with a speed of 10 m/s. The
angle of inclination is 30o
. The block goes a distance of _____ up the plane before
stopping. Assume g = 10 m/s2
6.0
To solve this problem, we can use the concepts of work and energy.
First, let's find the gravitational potential energy (GPE) of the block at its starting position. GPE is given by the equation GPE = mgh, where m is the mass of the block (2.0 kg), g is the acceleration due to gravity (10 m/s^2), and h is the vertical height above the reference point.
Since the block is projected up the inclined plane, the height above the reference point is given by h = d*sin(theta), where d is the distance up the plane and theta is the angle of inclination (30 degrees).
To find the vertical height, substitute the given values:
h = d*sin(30 degrees)
Now, let's find the kinetic energy (KE) of the block when it stops. KE is given by the equation KE = 0.5mv^2, where m is the mass of the block and v is its velocity.
Using the given values, we have:
KE = 0.5*(2.0 kg)*(10 m/s)^2
Since the block stops, all of its kinetic energy is converted into gravitational potential energy. Therefore, we can equate the two equations:
0.5*(2.0 kg)*(10 m/s)^2 = (2.0 kg)*(10 m/s^2)*(d*sin(30 degrees))
Simplifying the equation:
0.5*100 = 20*(d*sin(30 degrees))
50 = 20*(d*0.5)
50 = 10d
Solving for d:
d = 50/10
d = 5 meters
Therefore, the block goes a distance of 5 meters up the inclined plane before stopping.