One common application of conservation of energy in mechanics is to determine the speed of an object. although the simulation doesn’t give the skater's speed, you can calculate it because the skater's kinetic energy is known at any location on the track. consider again the case where the skater starts 7 m above the ground and skates down the track. what is the skater's speed when the skater is at the bottom of the track? express your answer numerically in meters per second to two significant figures.

11.72 m/s

Explanation:

Energy is conserved when there are no energy loses. So potential energy and kinetic energy must be the same at the beginning and at the end.

Mechanical energy is the addition of both energy types, which is conserved.

So E= P+ K,

E: Mechanical energy

P: Potential Energy

K: kinetic energy.

P= m*g*h

K= (m*v^2)/2

M:mass

G:gravity=9.8m/s^2

H:altitude=7m

V: Velocity.

So P+K is conserved,

P1+K1=P2+K2

At the beginning there's no movement, so V=0, then K1=0

At the end there's no altitude, so H=0, then P2=0.

For those reasons,

P1=K2

M*G*H= M*v^2*0.5. As M is in both sides, we can take it out, and replace gravity and altitude with the values we already have.

9,81(m/s^2)* 7m = v^2 *0.5

68.67 (m/s)^2 = v^2 *0.5

(68.67 (m/s)^2 :0.5) ^ (1/2)= |v|

137.34^0.5=|v|

11.72 m/s= |v|

The sign will depend on where are we considering the 0 so will it be positive velocity if it's increasing direction or negative if it is decreasing

11.71 m/s

Explanation:

In the absence of non-conservative forces, the mechanical energy which is the sum of kinetic energy and potential energy remains conserved. skaters kinetic energy down the track would be equal to potential energy at the top.

P.E. = m g h

where, m is the mass, g is the gravitational acceleration and h is the height.

K.E. = 0.5 m v²

where, v is the speed.

P.E. = K.E.

⇒m g h = 0.5 m v²

⇒v²= 2 g h

Substitute the values:

v² = 2 (9.8 m/s²) ( 7 m) = 137.2

⇒v = 11.71 m/s

a small object of mass 3.66 g and charge -17.6 µc is suspended motionless above the ground when immersed in a uniform electric field perpendicular to the ground. what are the magnitude and direction of the electric field?