Unit 5 Energy LTHW Packet AP Physics 1 Standard 5.2—Use work equations to find the energy input, forces or other aspect of an object’s motion 1) A block of mass m = 20kg is pulled by an applied force of 150N at an angle of 37° to the horizontal over a distance of 5m on a rough floor. The coefficient of kinetic friction between the block and floor is µ = 0.5. Determine the following quantities: a) The x and y components of the pulling force b) The normal force on the block by the floor c) The work done by the applied force during the pull: d) The work done by gravity during the pull: e) The work done by friction during the pull: f) The work done by the normal force during the pull: g) The change in kinetic energy of the block during the pull. h) After the block has been pulled the 5m, the pulling force is removed. How much farther will the block slide before coming to rest?
The Correct Answer and Explanation is:
Let’s go through each part of the problem step by step.
Given:
- Mass of block, m=20 kgm = 20 \, \text{kg}m=20kg
- Applied force, Fapplied=150 NF_{\text{applied}} = 150 \, \text{N}Fapplied=150N
- Angle of applied force, θ=37∘\theta = 37^\circθ=37∘
- Distance, d=5 md = 5 \, \text{m}d=5m
- Coefficient of kinetic friction, μ=0.5\mu = 0.5μ=0.5
a) The x and y components of the pulling force:
To break the applied force into its components, we use trigonometry.
- The horizontal component of the pulling force (Fapplied,xF_{\text{applied}, x}Fapplied,x) is: Fapplied,x=Fappliedcos(θ)=150 N×cos(37∘)=150×0.7986=119.79 NF_{\text{applied}, x} = F_{\text{applied}} \cos(\theta) = 150 \, \text{N} \times \cos(37^\circ) = 150 \times 0.7986 = 119.79 \, \text{N}Fapplied,x=Fappliedcos(θ)=150N×cos(37∘)=150×0.7986=119.79N
- The vertical component of the pulling force (Fapplied,yF_{\text{applied}, y}Fapplied,y) is: Fapplied,y=Fappliedsin(θ)=150 N×sin(37∘)=150×0.6018=90.27 NF_{\text{applied}, y} = F_{\text{applied}} \sin(\theta) = 150 \, \text{N} \times \sin(37^\circ) = 150 \times 0.6018 = 90.27 \, \text{N}Fapplied,y=Fappliedsin(θ)=150N×sin(37∘)=150×0.6018=90.27N
b) The normal force on the block by the floor:
The normal force is the force exerted by the floor on the block to counteract the vertical forces. It is affected by both the weight of the block and the vertical component of the applied force.
The weight of the block is:Fgravity=mg=20 kg×9.8 m/s2=196 NF_{\text{gravity}} = mg = 20 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 196 \, \text{N}Fgravity=mg=20kg×9.8m/s2=196N
The normal force is given by:Fnormal=Fgravity−Fapplied,y=196 N−90.27 N=105.73 NF_{\text{normal}} = F_{\text{gravity}} – F_{\text{applied}, y} = 196 \, \text{N} – 90.27 \, \text{N} = 105.73 \, \text{N}Fnormal=Fgravity−Fapplied,y=196N−90.27N=105.73N
c) The work done by the applied force during the pull:
Work is given by:W=F×d×cos(θ)W = F \times d \times \cos(\theta)W=F×d×cos(θ)
Since we are concerned with the horizontal component of the applied force, the work done by the applied force is:Wapplied=Fapplied,x×d=119.79 N×5 m=598.95 JW_{\text{applied}} = F_{\text{applied}, x} \times d = 119.79 \, \text{N} \times 5 \, \text{m} = 598.95 \, \text{J}Wapplied=Fapplied,x×d=119.79N×5m=598.95J
d) The work done by gravity during the pull:
Gravity acts vertically, and since the displacement is horizontal, the angle between the gravitational force and displacement is 90∘90^\circ90∘. The work done by gravity is zero:Wgravity=Fgravity×d×cos(90∘)=196 N×5 m×0=0 JW_{\text{gravity}} = F_{\text{gravity}} \times d \times \cos(90^\circ) = 196 \, \text{N} \times 5 \, \text{m} \times 0 = 0 \, \text{J}Wgravity=Fgravity×d×cos(90∘)=196N×5m×0=0J
e) The work done by friction during the pull:
The frictional force is given by:Ffriction=μFnormal=0.5×105.73 N=52.865 NF_{\text{friction}} = \mu F_{\text{normal}} = 0.5 \times 105.73 \, \text{N} = 52.865 \, \text{N}Ffriction=μFnormal=0.5×105.73N=52.865N
The work done by friction is:Wfriction=Ffriction×d×cos(180∘)=52.865 N×5 m×(−1)=−264.325 JW_{\text{friction}} = F_{\text{friction}} \times d \times \cos(180^\circ) = 52.865 \, \text{N} \times 5 \, \text{m} \times (-1) = -264.325 \, \text{J}Wfriction=Ffriction×d×cos(180∘)=52.865N×5m×(−1)=−264.325J
(The negative sign indicates that friction does work opposite to the direction of motion.)
f) The work done by the normal force during the pull:
The normal force is perpendicular to the displacement (horizontal), so the work done by the normal force is zero:Wnormal=Fnormal×d×cos(90∘)=105.73 N×5 m×0=0 JW_{\text{normal}} = F_{\text{normal}} \times d \times \cos(90^\circ) = 105.73 \, \text{N} \times 5 \, \text{m} \times 0 = 0 \, \text{J}Wnormal=Fnormal×d×cos(90∘)=105.73N×5m×0=0J
g) The change in kinetic energy of the block during the pull:
The work-energy theorem states that the net work done on an object is equal to its change in kinetic energy (ΔKE\Delta KEΔKE):ΔKE=Wapplied+Wgravity+Wfriction+Wnormal\Delta KE = W_{\text{applied}} + W_{\text{gravity}} + W_{\text{friction}} + W_{\text{normal}}ΔKE=Wapplied+Wgravity+Wfriction+Wnormal
Substituting the values we found:ΔKE=598.95 J+0 J−264.325 J+0 J=334.625 J\Delta KE = 598.95 \, \text{J} + 0 \, \text{J} – 264.325 \, \text{J} + 0 \, \text{J} = 334.625 \, \text{J}ΔKE=598.95J+0J−264.325J+0J=334.625J
Thus, the change in kinetic energy is 334.625 J334.625 \, \text{J}334.625J.
h) The distance the block slides before coming to rest:
When the pulling force is removed, only friction acts on the block. The work done by friction will equal the initial kinetic energy of the block. Using the work-energy theorem, we can find the distance the block travels before coming to rest.
The kinetic energy at the moment the pulling force is removed is 334.625 J334.625 \, \text{J}334.625J. The work done by friction is:Wfriction=Ffriction×dslide×(−1)W_{\text{friction}} = F_{\text{friction}} \times d_{\text{slide}} \times (-1)Wfriction=Ffriction×dslide×(−1)
Setting the work done by friction equal to the initial kinetic energy:334.625 J=52.865 N×dslide334.625 \, \text{J} = 52.865 \, \text{N} \times d_{\text{slide}}334.625J=52.865N×dslide
Solving for dslided_{\text{slide}}dslide:dslide=334.625 J52.865 N=6.33 md_{\text{slide}} = \frac{334.625 \, \text{J}}{52.865 \, \text{N}} = 6.33 \, \text{m}dslide=52.865N334.625J=6.33m
So, the block will slide about 6.33 meters before coming to rest.
