Question

A sled that has a mass of 8 kg is pulled at a 50 degree angle with a force of 20 N. The force of friction acting on the sled is 2.4 N. The free-body diagram shows the forces acting on the sled. A free body diagram with 4 force vectors. The first vector is pointing downward, labeled F Subscript g Baseline. The second vector is pointing up to the right at an angle of 50 degrees, labeled F Subscript p Baseline. The third vector is pointing upward, labeled F Subscript N Baseline. The fourth vector is pointing left, labeled F Subscript f Baseline. The up and down vectors are the same length. The right vector is longer than the left vector. What is the acceleration of the sled and the normal force acting on it, to the nearest tenth? a = 1.3 m/s2; FN = 63.1 N a = 1.6 m/s2; FN = 65.6 N a = 1.9 m/s2; FN = 93.7 N a = 2.2 m/s2; FN = 78.4 N

Answer 1

Answer:

1) a = 1.3 m/s2; FN = 63.1 N

Explanation:

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Answer 2

The acceleration of the sled will be 1.30 m/s². Force is defined as the product of mass and acceleration.

What is force?

Force is defined as the push or pulls applied to the body. Sometimes it is used to change the shape, size, and direction of the body.

Given data;

m(mass of sled)=8 kg

Θ is the inclination of force= 50°

Force of friction,f=2.4 N.

The applied force at the given angle is resolved into the two-component as;

$\rm F_h=F cos \theta \\\\ F_h= 20 cos 50 ^0 \\\\ F_h= 12.85 \ N$

$\rm F_v=F sin \theta \\\\ F_v=20 sin 50^0 \\\\ F_v=15.32 \ N$

The net vertical force is zero;

$\rm F_N=mg-Fsin50^0 \\\\ \rm F_N=8 \times 9.81 -15.32 \\\\ F_N=63.1 \ N$

From Newton's second law the net force as;

$\rm \sum F_{net}=ma \\\\ Fcos 60^0-f =ma \\\\ a=\frac{12.855-2.4}{8} \\\\a = 1.30 \ m/s^2$

Hence, the acceleration of the sled will be 1.30 m/s².

To learn more about the force refer to the link;

brainly.com/question/26115859

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