Physics, 04.03.2021 14:00 brashley

A 3.00 kg cart on a track is pulled by a string so that it accelerates at 2.00 m/s/s. The force of tension in the string is 10.0 N. What is the force of friction on the cart?

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Answer from: tangia

If the track is horizontal and the string is pulled horizontally, the friction on the cart would be $4.0\; \rm N$ .

Explanation:

Let denote the mass of this cart, and let denote the acceleration of this cart.

$m = 3.00\; \rm kg$ .

$a = 2.00\; \rm m \cdot s^{-2}$ .

Apply Newton's Second Law to find the net force on this cart.

$\begin{aligned}\text{Net Force} &= m \cdot a\\ &= 3.00\; \rm kg \times 2.00\; \rm m\cdot s^{-2}\\ &= 6.00\; \rm N\end{aligned}$ .

The following forces act upon this cart:

(downward) gravitational attraction from the earth,(upward) normal force from the track,(forward) tension from the string, and(backward) friction from the track.

Assume that the track is horizontal, and that the string was pulled horizontally. The normal force from the track would exactly balance the downward gravitational attraction from the earth. Hence, the $6.00\; \rm N$ net force on this cart would be equal (in size) to the size of the tension from the string ( $10.0\; \rm N$ ) minus the size of the friction from the track.

In other words:

$\begin{aligned}&\text{Size of Net Force}\\ &= \text{Size of Tension} - \text{size of friction}\end{aligned}$ .

$\begin{aligned}& 6.00\;\rm N = 10.0\; \rm N - (\text{size of friction})\end{aligned}$ .

$\text{size of friction} = 10.0\; \rm N - 6.00\; \rm N = 4.0\; \rm N$ .

Answer from: Quest

answer: 802.3 kilo joules of energy represents 191.75 kilo calorie of energy.

explanation:

energy released during combustion of one mole methane = 802.3 kilo joules

if 1 kilo joule = 0.239006 kilo calorie

then ,802.3 kilo joule will be = 802.3 $\times$ 0.239006 kilo calorie

802.3 kilo joule = 191.75 kilo calorie

hence, the energy in kilo calorie is 191.75 kilo calorie.

Answer from: Quest

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