Question

A 2.0 kg wood block is launched up a wooden ramp that is inclined at a 27 ∘ angle. The block's initial speed is 13 m/s . The coefficient of kinetic friction of wood on wood is μk=0.200. You may want to review (Page 131) . For help with math skills, you may want to review: Vector Components Part A What vertical height does the block reach above its starting point? Express your answer to two significant figures and include the appropriate units. View Available Hint(s) hh = nothing nothing Part B What speed does it have when it slides back down to its starting point? Express your answer to two significant figures and include the appropriate units. View Available Hint(s) vv = nothing nothing Provide Feedback

Answer 1

Answer:

a)

$6.2$ m

b)

$8.6$ ms⁻¹

Explanation:

$m$ = mass of the wood block = 2 kg

$v_{i}$ = initial speed of the block at the bottom = 13 m/s

$\theta$ = angle of incline = 27°

$\mu _{k}$ = Coefficient of kinetic friction = 0.2

$f _{k}$ = kinetic frictional force

$h$ = height of the incline gained

$L$ = length of the incline traveled

In triangle ABC

$Sin\theta =\frac{BC}{AB} =\frac{h}{L}$

$L = \frac{h}{Sin \theta}$

$N$ = Normal force on the block by incline surface

From the force diagram of the block, force equation perpendicular to the incline surface is given as

$N = mg Cos\theta$                            Eq-1

kinetic frictional force on the block is given as

$f_{k} = \mu _{k}N$

Using Eq-1

$f_{k} = \mu _{k} mg Cos\theta$

Work done by kinetic frictional force is given as

$W = f_{k} L$

Using conservation of energy

Kinetic energy at the bottom = work done by frictional force + potential energy at the top

$(0.5) m {v_{i}}^{2} = W + mgh$

$(0.5) m {v_{i}}^{2} = f_{k} L + mgh$

$(0.5) m {v_{i}}^{2} = \mu _{k} mg Cos\theta (\frac{h}{Sin \theta}) + mgh$

inserting the values

$(0.5) (2) (13)^{2} = (0.2) (2)(9.8) Cos27 (\frac{h}{Sin 27}) + (2)(9.8)h$

$h = 6.2$ m

b)

$v_{f}$ = final speed of the block after returning to starting point

Using conservation of energy

Kinetic energy at the bottom initially = work done by frictional force + Kinetic energy at the bottom finally

$(0.5) m {v_{i}}^{2} = W + (0.5) m {v_{f}}^{2}$

$(0.5) m {v_{i}}^{2} = f_{k} (2L) + (0.5) m {v_{f}}^{2}$

$(0.5) m {v_{i}}^{2} = \mu _{k} mg Cos\theta (\frac{2h}{Sin \theta}) + (0.5) m {v_{f}}^{2}$

inserting the values

$(0.5) (2) (13)^{2} = (0.2) (2)(9.8) Cos27 (\frac{2 (6.2))}{Sin 27}) + (0.5) (2) {v_{f}}^{2}$

$v_{f} = 8.6$ ms⁻¹