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Question4. NL11Q2 Points: 0/70 / 7
Amanda B. Reckenwyth is exerting a rightward force of 139.0 N in order to drag a 20.3−kg20.3-\mathrm{kg} box of books across the floor at a constant speed of 0.40 m/s0.40 \mathrm{~m} / \mathrm{s}. Determine the ... a. ... force of gravity. Fgrav Info Attempts: 0/0 Submit b. ... normal force. Fnorm Info Attempts: 0/00 / 0 Submit c. .- applied force. Fapp Info Attempts: 0/00 Submit d. .. friction force. Ffrict Info Attempts: 0/00 Submit

Studdy Solution

STEP 1

1. The box is moving at a constant speed, indicating that the net force is zero.
2. The force of gravity acts downward on the box.
3. The normal force acts perpendicular to the surface, balancing the force of gravity.
4. The applied force is given as 139.0 N 139.0 \, \text{N} to the right.
5. The friction force opposes the motion and is equal in magnitude to the applied force due to constant speed.

STEP 2

1. Calculate the force of gravity.
2. Determine the normal force.
3. Confirm the applied force.
4. Calculate the friction force.

STEP 3

Calculate the force of gravity using the formula:
Fgrav=m⋅g F_{\text{grav}} = m \cdot g
where m=20.3 kg m = 20.3 \, \text{kg} and g=9.8 m/s2 g = 9.8 \, \text{m/s}^2 .
Fgrav=20.3 kg×9.8 m/s2 F_{\text{grav}} = 20.3 \, \text{kg} \times 9.8 \, \text{m/s}^2 =198.94 N = 198.94 \, \text{N}

STEP 4

Determine the normal force. Since the box is on a horizontal surface and there is no vertical acceleration, the normal force equals the gravitational force:
Fnorm=Fgrav=198.94 N F_{\text{norm}} = F_{\text{grav}} = 198.94 \, \text{N}

STEP 5

Confirm the applied force. The applied force is given as:
Fapp=139.0 N F_{\text{app}} = 139.0 \, \text{N}

STEP 6

Calculate the friction force. Since the box moves at a constant speed, the friction force equals the applied force:
Ffrict=Fapp=139.0 N F_{\text{frict}} = F_{\text{app}} = 139.0 \, \text{N}
The forces are: a. Force of gravity: 198.94 N \boxed{198.94 \, \text{N}} b. Normal force: 198.94 N \boxed{198.94 \, \text{N}} c. Applied force: 139.0 N \boxed{139.0 \, \text{N}} d. Friction force: 139.0 N \boxed{139.0 \, \text{N}}

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