Question

A child on a sled slides (starting from rest) down an icy slope that makes an angle of 15◦ with the horizontal. After sliding 20 m down the slope, the child enters a flat, slushy region, where she slides for 2.0 s with a constant negative acceleration of −1.5 m/s2 with respect to her direction of motion. She then slides up another icy slope that makes a 20◦ angle with the horizontal.
A) How fast was the child going when she reached the bottom of the first slope? How long did it take her to get there?B) How long was the flat stretch at the bottom?C) How fast was the child going as she started up the second slope?D) How far up the second slope did she slide?

Answer 1

Answer:

A) v₁ = 10.1 m/s t₁= 4.0 s

B) x₂= 17.2 m

C) v₂=7.1 m/s

D) x₂=7.5 m

Explanation:

A)

• Assuming no friction, total mechanical energy must keep constant, so the following is always true:

       $\Delta K + \Delta U = (K_{f} - K_{o}) +( U_{f} - U_{o}) = 0 (1)$

• Choosing the ground level as our zero reference level, Uf =0.
• Since the child starts from rest, K₀ = 0.
• From (1), ΔU becomes:
• $\Delta U = 0- m*g*h = -m*g*h (2)$  
• In the same way, ΔK becomes:
• $\Delta K = \frac{1}{2}*m*v_{1}^{2} (3)$      
• Replacing (2) and (3) in (1), and simplifying, we get:

       $\frac{1}{2}*v_{1}^{2} = g*h (4)$

• In order to find v₁, we need first to find h, the height of the slide.
• From the definition of sine of an angle, taking the slide as a right triangle, we can find the height h, knowing the distance that the child slides down the slope, x₁, as follows:

       $h = x_{1} * sin \theta_{1} = 20.0 m * sin 15 = 5.2 m (5)$

       Replacing (5) in (4) and solving for v₁, we get:

      $v_{1} = \sqrt{2*g*h} = \sqrt{2*9.8m/s2*5.2m} = 10.1 m/s (6)$

• As this speed is achieved when all the energy is kinetic, i.e. at the bottom of the first slide, this is the answer we were looking for.
• Now, in order to finish A) we need to find the time that the child used to reach to that point, since she started to slide at the its top.
• We can do this in more than one way, but a very simple one is using kinematic equations.
• If we assume that the acceleration is constant (which is true due the child is only accelerated by gravity), we can use the following equation:

       $v_{1}^{2} - v_{o}^{2} = 2*a* x_{1} (7)$

• Since v₀ = 0 (the child starts from rest) we can solve for a:

       $a = \frac{v_{1}^{2}}{2*x_{1} } = \frac{(10.1m/s)^{2}}{2* 20.0m} = 2.6 m/s2 (8)$

• Since v₀ = 0, applying the definition of acceleration, if we choose t₀=0, we can find t as follows:

       $t_{1} =\frac{v_{1} }{a} =\frac{10.1m/s}{2.6m/s2} = 4.0 s (9)$

B)

• Since we know the initial speed for this part, the acceleration, and the time, we can use the kinematic equation for displacement, as follows:

       $x_{2} = v_{1} * t_{2} + \frac{1}{2} *a_{2}*t_{2}^{2} (10)$

• Replacing the values of v₁ = 10.1 m/s, t₂= 2.0s and a₂=-1.5m/s2 in (10):

       $x_{2} = 10.1m/s * 2.0s + \frac{1}{2} *(-1.5m/s2)*(2.0s)^{2} = 17.2 m (11)$

C)  

• From (6) and (8), applying the definition for acceleration, we can find the speed of the child whem she started up the second slope, as follows:

       $v_{2} = v_{1} + a_{2} *t_{2} = 10.1m/s - 1.5m/s2*2.0s = 7.1 m/s (12)$

D)

• Assuming no friction, all the kinetic energy when she started to go up the second slope, becomes gravitational potential energy when she reaches to the maximum height (her speed becomes zero at that point), so we can write the following equation:

       $\frac{1}{2}*v_{2}^{2} = g*h_{2} (13)$

• Replacing from (12) in (13), we can solve for h₂:

       $h_{2} =\frac{v_{2} ^{2}}{2*g} = \frac{(7.1m/s) ^{2}}{2*9.8m/s2} = 2.57 m (14)$

• Since we know that the slide makes an angle of 20º with the horizontal, we can find the distance traveled up the slope applying the definition of sine of an angle, as follows:

       $x_{3} = \frac{h_{2} }{sin 20} = \frac{2.57m}{0.342} = 7.5 m (15)$