A crate of mass \(m\) is released from rest at the top of a stationary ramp of height \(h\), whose length along the curve is \(s\), as shown. Assuming that the ramp is frictionless, in calculating the speed of the crate \(v\) at the bottom of the ramp, one can assume: (A) \(h=v t\) (B) \(m g h=1 / 2 m v^2\) (C) \(s=1 / 2 a t^2\) (D) \(v^2=2 a s\) (E) \(h=1 / 2 g t^2\)

Since there is no friction, mechanical energy is conserved. Initially, the crate has potential energy \(E_p = mgh\). At the bottom of the ramp, the crate has kinetic energy \(E_k = \frac{1}{2}mv^2\). The conservation of energy states that the total energy at the beginning and the end must remain the same. Therefore, we can write the equation as: \(mgh=\frac{1}{2}mv^2\)

Now, let's analyze each given option and see whether they match our energy conservation equation or not. (A) \(h=vt\) - This option refers to the height reached by the crate when it's moving at a constant speed. It is unrelated to our problem of finding the speed at the bottom of the ramp. (B) \(mgh=\frac{1}{2}mv^2\) - This option is the same as our energy conservation equation and, therefore, correct. (C) \(s=\frac{1}{2}at^2\) - This kinematic equation refers to the distance covered with a constant acceleration and initial velocity of zero. It is true, but doesn't directly give us the speed of the crate at the bottom of the ramp. (D) \(v^2=2as\) - This kinematic equation relates final speed, initial speed, acceleration, and distance covered. The acceleration isn't constant, so it isn't applicable here. (E) \(h=\frac{1}{2}gt^2\) - This equation refers to the height reached in a vertical motion due to gravity after a time interval \(t\). It is unrelated to our problem. So, the correct expression for calculating the speed of the crate \(v\) at the bottom of the ramp is: (B) \(mgh=\frac{1}{2}mv^2\).