Answer:
(a) The acceleration is 1.98 m/s²
(b) The velocity of the piano is 3.98 i m/s
Step-by-step explanation:
In order to answer part (a) you have to apply Newton's Second Law which is:
∑F=mA
where F is all the forces, m is the mass and A is the acceleration.
You have to perfom a Free Body Diagram for the piano (see the attachment)
Applying Newton's Second Law:
x-axis: F - mgSin(20°) =mA
Notice that mgSin(20°) is the x component of the weight
Replacing the given values:
1600 - (300)(9.8)Sin(20°) = 300A
594.5 = 300A
Dividing by 300 both sides:
A=594.5/300
A=1.98 m/s²
For part (b) you have to apply the Work-Energy Theorem which is:
Work of all non-conservative forces = ΔEm
where ΔEm is the variation of mechanical energy in two points a and b
The work is represented by:
W=

Notice that the only force producing work is the force of 1600 N because is the only non-conservative force in the direction of the displacement.
In this case, ds: dx i
F= 1600 i
a= 0 m and b =4.0 m
So, you have to calculate the dot product between vectors F and ds which is:
|F||ds|Cos(0°)=1600dx
The angle between them is 0° because they are parallels and in the same direction.
Calculating the work:
W=
evaluated in x=0 and x=4
W=1600(4)-1600(0)=6400 J
Now, ΔEm=ΔEk+ΔEp
where ΔEk is the variation of kinetic enery and ΔEp is the variation of potential energy
In this case ΔEp=mgh, where h is the height of the ramp (4Sin20)
ΔEk between a and b = Ekb-Eka
Ekb-Eka=6400
Eka=0 because the piano starts from rest (Va=0)
Ekb=6400


Solving for Vb:
Vb= 3.98 m/s
So, the velocity vector is:
Vb= 3.98 i m/s